ESAT Physics Mock 2

1. Calculate the useful mechanical power output of the motor:2. Calculate the rate at which energy is wasted (dissipated as thermal energy, sound, etc.):3. Convert the time from minutes to seconds:4. Calculate the total energy wasted in this time:Therefore, the correct option is C.

Let the parent atom have $\displaystyle Z$ protons, $\displaystyle N_n$ neutrons, and $\displaystyle E$ electrons. The total number of particles is $\displaystyle N = Z + N_n + E$ . Since the atom is neutral, the number of electrons equals the number of protons, so $\displaystyle E = Z$ . Thus, $\displaystyle N = Z + N_n + Z = 2Z + N_n$ .

Alpha decay is the emission of an alpha particle, which is a helium nucleus ( $\displaystyle ^4_2\text{He}$ ). An alpha particle contains 2 protons and 2 neutrons.

1. Change in the nucleus: The parent nucleus loses 2 protons and 2 neutrons. The number of protons in the daughter nucleus is $\displaystyle Z' = Z - 2$ . The number of neutrons is $\displaystyle N'_n = N_n - 2$ .

2. Change in electrons: The parent atom had $\displaystyle Z$ electrons. The daughter product is a neutral atom, so its number of electrons must equal its new number of protons. The daughter atom must have $\displaystyle E' = Z' = Z - 2$ electrons. This means the atom has lost $\displaystyle Z - (Z-2) = 2$ electrons.

3. Total change in particles: The total number of particles lost from the atom is the sum of protons, neutrons, and electrons lost.
Total loss = (2 protons) + (2 neutrons) + (2 electrons) = 6 particles.

Therefore, the total number of particles in the new daughter atom is $\displaystyle N - 6$ .

First, we extract the amplitude ( $\displaystyle A$ ) and wavelength ( $\displaystyle \lambda$ ) from the graph.
- The amplitude is the maximum displacement, so $\displaystyle A = 2.5\,\text{cm} = 0.025\,\text{m}$ .
- The wavelength is the length of one complete cycle, so $\displaystyle \lambda = 0.5\,\text{m}$ .

Next, we use the wave speed equation, $\displaystyle v = f\lambda$ , to find the frequency $\displaystyle f$ . The wave speed is given as $\displaystyle v = 10\,\text{m}\,\text{s}^{-1}$ .Now, we calculate the average speed of a particle on the string. In one complete oscillation (one period, $\displaystyle T$ ), the particle moves from its equilibrium position up to the crest, back to equilibrium, down to the trough, and back to equilibrium. The total distance travelled is four times the amplitude.

Total distance in one cycle, $\displaystyle d = 4A = 4 \times 0.025\,\text{m} = 0.1\,\text{m}$ .

The time for one cycle is the period, $\displaystyle T = \dfrac{1}{f} = \dfrac{1}{20}\,\text{s}$ .

The average speed is total distance divided by total time.Substituting the values we found:

Since the power output $\displaystyle P$ is constant and there are no resistive forces, the work done on the vehicle up to time $\displaystyle t$ equals its kinetic energy: $\displaystyle Pt = \dfrac{1}{2}m v^2 \implies v(t) = \sqrt{\dfrac{2Pt}{m}} = \left(\dfrac{2P}{m}\right)^{1/2} t^{1/2}$ The acceleration $\displaystyle a(t)$ is the time derivative of velocity: $\displaystyle a(t) = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{1}{2}\left(\dfrac{2P}{m}\right)^{1/2} t^{-1/2} = \sqrt{\dfrac{P}{2m}}\, t^{-1/2}$ To determine how the magnitude of acceleration changes over time:
1. Since $\displaystyle a(t) \propto t^{-1/2}$ , the magnitude of $\displaystyle a(t)$ decreases as $\displaystyle t$ increases.
2. The rate of change of acceleration is given by the derivative: $\displaystyle \dfrac{\mathrm{d}a}{\mathrm{d}t} = -\dfrac{1}{2}\sqrt{\dfrac{P}{2m}}\, t^{-3/2} < 0$ 3. The magnitude of this rate of change, $\displaystyle \left|\dfrac{\mathrm{d}a}{\mathrm{d}t}\right| \propto t^{-3/2}$ , decreases as $\displaystyle t$ increases (or equivalently, $\displaystyle \dfrac{\mathrm{d}^2a}{\mathrm{d}t^2} > 0$ ).

Therefore, the acceleration is decreasing, and the rate at which it decreases is also decreasing (it is decreasing at a decreasing rate).

Relative to a proton, the nucleus has mass-to-charge ratio $\displaystyle A/Z=2.5$ , where $\displaystyle A$ is protons plus neutrons and $\displaystyle Z$ is protons. For option B, $\displaystyle A=2+3=5$ and $\displaystyle Z=2$ , so $\displaystyle A/Z=5/2=2.5$ . Therefore the correct option is B.

The extra return time is $\displaystyle 70-30=40\,\mu\mathrm{s}$ . It corresponds to travelling through the layer twice. Hence $\displaystyle d=\dfrac{1500\times40\times10^{-6}}{2}=0.030\,\mathrm{m}=3.0\,\mathrm{cm}.$

Alpha particles are positive and are attracted towards the lower negative plate. Beta-minus particles are negative and are attracted towards the upper positive plate. Uncharged gamma radiation is not deflected.

Let $\displaystyle N_0$ be the initial number of atoms of the parent isotope P at time $\displaystyle t=0$ . At this time, the number of daughter atoms D is 0. The total number of atoms, $\displaystyle N_P + N_D$ , is always conserved and equal to $\displaystyle N_0$ .

At time $\displaystyle t$ , we are given that the ratio $\displaystyle N_D : N_P = 3:1$ , which means $\displaystyle N_D = 3N_P$ .

Using the conservation of the total number of atoms:Substituting $\displaystyle N_D = 3N_P$ into this equation:This means that at time $\displaystyle t$ , one quarter of the original parent atoms remain. The fraction of parent atoms remaining after $\displaystyle n$ half-lives is $\displaystyle (\dfrac{1}{2})^n$ . For the fraction to be $\displaystyle \dfrac{1}{4}$ , we have:So, $\displaystyle n=2$ . Two half-lives have passed by time $\displaystyle t$ . Let $\displaystyle T_{1/2}$ be the half-life. Then $\displaystyle t = 2T_{1/2}$ .

The question asks for the ratio at time $\displaystyle 2t$ . This corresponds to a total time of $\displaystyle 2 \times (2T_{1/2}) = 4T_{1/2}$ , which is 4 half-lives from the start.

After 4 half-lives, the number of parent atoms remaining, $\displaystyle N_P'$ , will be:The number of daughter atoms, $\displaystyle N_D'$ , will be the rest:The new ratio of D to P is:So the ratio is $\displaystyle 15:1$ .

The energy required to heat the water is given by $\displaystyle Q = mc\Delta T$ , where $\displaystyle m$ is the mass of water and $\displaystyle c$ is its specific heat capacity. The rate of energy transfer is power, $\displaystyle P = \dfrac{dQ}{dt}$ . Therefore, $\displaystyle P = mc \dfrac{d T}{dt}$ . Since $\displaystyle m$ and $\displaystyle c$ are constant, the power $\displaystyle P$ is directly proportional to the rate of change of temperature, $\displaystyle \dfrac{dT}{dt}$ , which is the gradient of the temperature-time graph.

The electrical power dissipated by the heating element is given by $\displaystyle P = \dfrac{V^2}{R_{total}}$ , where $\displaystyle V$ is the constant supply voltage.

Let the resistance of one resistor segment be $\displaystyle r$ . Initially, there are three resistors in series, so the total resistance is:The initial power is $\displaystyle P_{initial} = \dfrac{V^2}{3r}$ . The gradient of Graph X is proportional to this power.

When one resistor fails and becomes a short circuit, its resistance becomes zero. The new total resistance is the sum of the remaining two resistors:The new power is $\displaystyle P_{new} = \dfrac{V^2}{2r}$ .

The ratio of the new gradient to the old gradient is equal to the ratio of the new power to the old power:The new gradient is $\displaystyle 1.5$ times the original gradient. The heating process is faster. Graph X shows the temperature rising from 20 °C to 80 °C (a change of 60 °C) in 60 s. The new time taken for the same temperature change will be $\displaystyle \dfrac{60 \text{ s}}{1.5} = 40$ s. The new graph must be a straight line starting at (0, 20) and passing through (40, 80). This corresponds to line P.

Let $\displaystyle Q$ be the total thermal energy transferred. Since the mass of the liquid and the temperature change are identical in both cases, $\displaystyle Q$ is the same for both heaters.

The relationship between energy $\displaystyle Q$ , power $\displaystyle P$ , and time $\displaystyle t$ is $\displaystyle Q = P \times t$ , which can be rearranged to $\displaystyle t = Q/P$ .

Let $\displaystyle P_1 = 150\text{ W}$ and $\displaystyle P_2 = 100\text{ W}$ . The times taken are $\displaystyle t_1 = Q/P_1$ and $\displaystyle t_2 = Q/P_2$ .

The heater with the higher power will be faster, so $\displaystyle t_1 < t_2$ . The problem states that the time difference is 60 s.Substituting the expressions for $\displaystyle t_1$ and $\displaystyle t_2$ :Factor out $\displaystyle Q$ :To subtract the fractions, find a common denominator, which is 300:Solving for $\displaystyle Q$ :Converting to kilojoules:

The field around P is anticlockwise, so at M, to its right, the field points upwards. Reversing Q makes its field clockwise; at M, to its left, that field also points upwards. The two equal fields add to $\displaystyle 2B$ .

1. Current distribution: Because the frame is made of uniform wire, the two parallel paths from $\displaystyle P$ to $\displaystyle R$ ( $\displaystyle P \to Q \to R$ and $\displaystyle P \to S \to R$ ) have identical lengths and resistances ( $\displaystyle 2L$ ). Thus, a current of $\displaystyle \dfrac{I}{2}$ flows along each branch.

2. **Forces on vertical segments ( $\displaystyle PQ$ and $\displaystyle SR$ ):**
- Along $\displaystyle PQ$ , current flows upwards from $\displaystyle P$ to $\displaystyle Q$ .
- Along $\displaystyle SR$ , current flows upwards from $\displaystyle S$ to $\displaystyle R$ .
- In both cases, the current is parallel to the magnetic field (which points towards the top of the page). Since $\displaystyle \theta = 0^\circ$ , $\displaystyle \sin 0^\circ = 0$ , so the magnetic force on both $\displaystyle PQ$ and $\displaystyle SR$ is zero ( $\displaystyle F = 0$ ).

3. **Forces on horizontal segments ( $\displaystyle QR$ and $\displaystyle PS$ ):**
- Along $\displaystyle QR$ , current of $\displaystyle \dfrac{I}{2}$ flows from left to right ( $\displaystyle Q \to R$ ).
- Along $\displaystyle PS$ , current of $\displaystyle \dfrac{I}{2}$ flows from left to right ( $\displaystyle P \to S$ ).
- Using Fleming's Left-Hand Rule:
- Field (Index finger): towards top of page
- Current (Second finger): towards right of page
- Force (Thumb): directed perpendicularly out of the plane of the page.
- Both segments $\displaystyle QR$ and $\displaystyle PS$ experience equal forces of magnitude $\displaystyle F = B\left(\dfrac{I}{2}\right)L$ directed out of the plane of the page.

4. Net Force and Torque:
- The total force is non-zero and directed perpendicularly out of the page: $\displaystyle F_{\text{net}} = BIL$ .
- Because the two forces act symmetrically and in the same direction at equal distances from the centre of mass, the net torque about any central axis is zero.
- Therefore, the frame translates perpendicularly out of the plane of the page without rotating.

We first calculate the total energy supplied by the heater over the $\displaystyle 200\text{ s}$ period:Next, we determine the energy stored in the block (increasing its internal energy) using $\displaystyle E = mc\Delta T$ :By conservation of energy, the difference between the energy supplied and the energy stored is the energy lost to the surroundings:The average rate of energy loss is therefore:

The nearby north pole induces a south pole at X and a north pole at Y. Outside a north pole the field points away, so the compass north end points away from Y. Soft iron is useful because its induced magnetisation largely disappears when the applied field is removed.

1. Volume of one cylindrical rod: $\displaystyle V_{\text{rod}} = \pi r^2 L = \pi (1.0\text{ cm})^2 \times 10.0\text{ cm} = 10.0\pi\text{ cm}^3$ 2. Internal volume of the tube: $\displaystyle V_{\text{tube}} = \pi R^2 L = \pi (3.0\text{ cm})^2 \times 10.0\text{ cm} = 90.0\pi\text{ cm}^3$ 3. Volume of all seven rods: $\displaystyle V_{\text{rods}} = 7 \times 10.0\pi\text{ cm}^3 = 70.0\pi\text{ cm}^3$ 4. Volume occupied by the resin: $\displaystyle V_{\text{resin}} = V_{\text{tube}} - V_{\text{rods}} = 90.0\pi\text{ cm}^3 - 70.0\pi\text{ cm}^3 = 20.0\pi\text{ cm}^3$ 5. Mass of each component:
- Metal X (1 central rod): $\displaystyle m_X = 10.0\pi\text{ cm}^3 \times 8.0\text{ g cm}^{-3} = 80.0\pi\text{ g}$ - Metal Y (6 outer rods): $\displaystyle m_Y = 6 \times (10.0\pi\text{ cm}^3 \times 3.0\text{ g cm}^{-3}) = 180.0\pi\text{ g}$ - Resin: $\displaystyle m_{\text{resin}} = 20.0\pi\text{ cm}^3 \times 1.5\text{ g cm}^{-3} = 30.0\pi\text{ g}$ 6. Total mass of the contents: $\displaystyle m_{\text{total}} = 80.0\pi + 180.0\pi + 30.0\pi = 290\pi\text{ g}$

The nucleus is initially at rest, so the initial total momentum of the system is zero. By the principle of conservation of linear momentum, the total momentum after the decay must also be zero.

Let the alpha particle have momentum $\displaystyle p_\alpha$ and the daughter nucleus have momentum $\displaystyle p_d$ . Conservation of momentum requires:This implies that their momenta are equal in magnitude and opposite in direction: $\displaystyle |p_d| = |p_\alpha|$ . Let this magnitude be $\displaystyle p$ .

The kinetic energy (KE) of a particle with mass $\displaystyle m$ and momentum $\displaystyle p$ is given by the relationship $\displaystyle KE = \dfrac{p^2}{2m}$ .

The kinetic energy of the alpha particle is given as $\displaystyle K$ . So, we have:The kinetic energy of the daughter nucleus, $\displaystyle KE_d$ , is:From the equation for the alpha particle's kinetic energy, we can express $\displaystyle p^2$ as:Now, substitute this expression for $\displaystyle p^2$ into the equation for the daughter nucleus's kinetic energy:Simplifying this expression gives the final result:

Let the angle of a ray be $\displaystyle \phi$ and the angle of the mirror normal be $\displaystyle \alpha$ . The law of reflection in a 2D plane can be written as:For a small change in angles, the relationship is $\displaystyle \Delta \phi_{\text{out}} = 2\Delta \alpha - \Delta \phi_{\text{in}}$ .

1. **First Mirror ( $\displaystyle M_1$ ):** The incident beam is fixed ( $\displaystyle \Delta \phi_{\text{in},1} = 0$ ). $\displaystyle M_1$ rotates by $\displaystyle \theta$ clockwise ( $\displaystyle \Delta \alpha_1 = \theta$ ).The ray leaving $\displaystyle M_1$ rotates by $\displaystyle 2\theta$ clockwise.

2. **Second Mirror ( $\displaystyle M_2$ ):** This ray becomes the incident ray for $\displaystyle M_2$ , so $\displaystyle \Delta \phi_{\text{in},2} = 2\theta$ . We want the final direction to be unchanged ( $\displaystyle \Delta \phi_{\text{out},2} = 0$ ). Let $\displaystyle M_2$ rotate by $\displaystyle \Delta \alpha_2$ .Therefore, $\displaystyle M_2$ must rotate by $\displaystyle \theta$ in the same direction (clockwise).

First, determine the average density of the block, $\displaystyle \rho_{\text{avg}}$ . For an object floating in a fluid, the buoyant force equals its weight. This implies that the average density of the object is equal to the fraction of its volume submerged multiplied by the fluid density. $\displaystyle W = F_B$ $\displaystyle m_{\text{total}} g = \rho_{\text{fluid}} V_{\text{submerged}} g$ $\displaystyle (\rho_{\text{avg}} V_{\text{total}}) g = \rho_{\text{fluid}} (f V_{\text{total}}) g$ where $\displaystyle f$ is the fraction submerged. This simplifies to $\displaystyle \rho_{\text{avg}} = f \times \rho_{\text{fluid}}$ .

Given $\displaystyle f = \dfrac{4}{5}$ and $\displaystyle \rho_{\text{water}} = 1000\, \text{kg}\,\text{m}^{-3}$ : $\displaystyle \rho_{\text{avg}} = \dfrac{4}{5} \times 1000 = 800\, \text{kg}\,\text{m}^{-3}$ Now, we can find the total mass of the block: $\displaystyle m_{\text{total}} = \rho_{\text{avg}} V_{\text{total}} = 800\, \text{kg}\,\text{m}^{-3} \times 1.0\, \text{m}^3 = 800\, \text{kg}$ .

Next, we need to find the volumes of the constituent materials, $\displaystyle V_A$ and $\displaystyle V_B$ . We have a system of two equations:
1. Total volume: $\displaystyle V_A + V_B = V_{\text{total}} = 1.0\, \text{m}^3$ 2. Total mass: $\displaystyle m_A + m_B = m_{\text{total}}$ Substituting densities and volumes: $\displaystyle \rho_A V_A + \rho_B V_B = m_{\text{total}}$ $\displaystyle 200 V_A + 1200 V_B = 800$ From equation (1), $\displaystyle V_A = 1.0 - V_B$ . Substitute this into the second equation: $\displaystyle 200(1.0 - V_B) + 1200 V_B = 800$ $\displaystyle 200 - 200 V_B + 1200 V_B = 800$ $\displaystyle 1000 V_B = 800 - 200$ $\displaystyle 1000 V_B = 600$ $\displaystyle V_B = \dfrac{600}{1000} = 0.6\, \text{m}^3$ Finally, calculate the mass of material B: $\displaystyle m_B = \rho_B V_B = 1200\, \text{kg}\,\text{m}^{-3} \times 0.6\, \text{m}^3 = 720\, \text{kg}$ .

The definitions for kinetic energy (KE) and momentum (p) are:We can find a relationship between KE, p, and speed v. One way is to divide the KE equation by the momentum equation:Cancelling m and one factor of v gives:Rearranging for speed v:From the graph, at point X, $\displaystyle KE = 12\, \text{J}$ and $\displaystyle p = 6.0\, \text{Ns}$ . Substituting these values:

Halving the primary turns doubles the secondary voltage. With the same load resistance, secondary current also doubles, so output power quadruples. Input power therefore quadruples; since primary voltage is fixed, primary current increases by a factor of four.

1. Identify all vertical forces acting on the probe:
- Upward force: $\displaystyle \text{upthrust} = 540\text{ N}$ - Downward forces: $\displaystyle \text{weight} = 400\text{ N}$ and $\displaystyle \text{drag} = 60\text{ N}$ 2. Calculate the resultant vertical force: $\displaystyle F_{\text{net}} = 540\text{ N} - (400\text{ N} + 60\text{ N}) = 540\text{ N} - 460\text{ N} = 80\text{ N} \text{ (upwards)}$ 3. Use Newton's second law ( $\displaystyle F = ma$ ) to find the acceleration: $\displaystyle a = \dfrac{F_{\text{net}}}{m} = \dfrac{80\text{ N}}{40\text{ kg}} = 2.0\text{ m s}^{-2} \text{ (upwards)}$ Therefore, row A is correct.

Particles have a range of kinetic energies. Those with enough energy can escape from the surface. Preferential loss of these higher-energy particles reduces the average kinetic energy, and hence the temperature, of the remaining water.

Let $\displaystyle R(t)$ be the count rate due to the radioactive sample alone, and let $\displaystyle B$ be the constant background count rate. The total recorded count rate is $\displaystyle C(t) = R(t) + B$ .

In each interval of $\displaystyle \Delta t = 40\text{ minutes}$ , the sample's count rate decays by a constant factor $\displaystyle k$ , so $\displaystyle R(40) = k R(0)$ and $\displaystyle R(80) = k^2 R(0)$ .

Subtracting consecutive total count rates eliminates $\displaystyle B$ : $\displaystyle C(0) - C(40) = R(0)(1 - k) = 420 - 120 = 300$ $\displaystyle C(40) - C(80) = k R(0)(1 - k) = 120 - 45 = 75$ Dividing the second equation by the first gives: $\displaystyle k = \dfrac{75}{300} = \dfrac{1}{4}$ Since $\displaystyle k = (1/2)^2 = 1/4$ , the sample undergoes 2 half-lives in 40 minutes. Therefore: $\displaystyle T_{1/2} = \dfrac{40\text{ min}}{2} = 20\text{ minutes}$ Substituting $\displaystyle k = 1/4$ into $\displaystyle R(0)(1 - k) = 300$ : $\displaystyle R(0)\left(1 - \dfrac{1}{4}\right) = 300 \implies \dfrac{3}{4} R(0) = 300 \implies R(0) = 400\text{ counts per min}$ The background count rate is then: $\displaystyle B = C(0) - R(0) = 420 - 400 = 20\text{ counts per min}$ Thus, row B is correct.

In $\displaystyle 0.120\ \text{s}$ , the sound travelsThe outward journey is $\displaystyle d$ . During the round-trip time the bat moves $\displaystyle 5.0(0.120)=0.60\ \text{m}$ closer, so the return journey is $\displaystyle d-0.60$ . ThereforeThus $\displaystyle 2d=41.4$ and $\displaystyle d=20.7\ \text{m}$ . The correct option is C.

Let $\displaystyle p$ be the number of protons, $\displaystyle n$ the number of neutrons, and $\displaystyle e$ the number of electrons in the ion $\displaystyle \text{P}^{+}$ .

Since $\displaystyle \text{P}^{+}$ has a relative charge of $\displaystyle +1$ , it has $\displaystyle e = p - 1$ .
The total number of particles in $\displaystyle \text{P}^{+}$ is: $\displaystyle p + n + e = p + n + (p - 1) = 2p + n - 1 = 23 \implies 2p + n = 24$ Thus, the number of neutrons in $\displaystyle \text{P}^{+}$ is $\displaystyle n = 24 - 2p$ .

For any candidate atom or ion to be of the same element, it must have the same number of protons, $\displaystyle p$ .
If a candidate has total particle count $\displaystyle N$ , relative charge $\displaystyle q$ , and neutron count $\displaystyle n'$ , then its electron count is $\displaystyle e' = p - q$ , so: $\displaystyle N = p + n' + e' = p + n' + (p - q) = 2p + n' - q \implies n' = N - 2p + q$ The difference in neutron count between the candidate and $\displaystyle \text{P}^{+}$ is: $\displaystyle n' - n = (N - 2p + q) - (24 - 2p) = N + q - 24$ A species is a *different isotope* if $\displaystyle n' e n$ (i.e. $\displaystyle N + q e 24$ ), and the *same isotope* if $\displaystyle n' = n$ (i.e. $\displaystyle N + q = 24$ ):
- W: $\displaystyle N + q = 24 + 0 = 24 \implies n' = n$ (same isotope)
- X: $\displaystyle N + q = 25 + 0 = 25 \implies n' = n + 1 e n$ (different isotope)
- Y: $\displaystyle N + q = 26 + (-1) = 25 \implies n' = n + 1 e n$ (different isotope)
- Z: $\displaystyle N + q = 22 + 1 = 23 \implies n' = n - 1 e n$ (different isotope)

Therefore, X, Y, and Z could be of a different isotope of the same element as $\displaystyle \text{P}^{+}$ .

The wave speed isThe frequency is $\displaystyle 2.0\ \text{Hz}$ . Using $\displaystyle v=f\lambda$ ,The correct option is D.