Work, Energy, Power and Collisions Questions and Answers

Answer: C

The ball loses 15 per cent of its energy.

Its energy just before impact equals its potential energy at the starting height, and its energy just after equals the potential energy at the height it rebounds to, so energy is proportional to height.

\[ \begin{aligned} \text{loss} &= \frac{mg(20 - 17)}{mg \times 20} \times 100 \\ &= \frac{3}{20} \times 100 = 15 \end{aligned} \]

Dividing the 3 m loss by the rebound height of 17 m instead of the original 20 m is the usual slip; 7.5 and 5 are half and a third of the right value.

Answer: E

The force constant is \(1 \times 10^{5}\) Nm^-1.

On a frictionless surface the block's kinetic energy is stored entirely in the spring:

\[ \begin{aligned} \tfrac{1}{2}mv^{2} &= \tfrac{1}{2}kx^{2} \\ k &= \frac{mv^{2}}{x^{2}} = \frac{10 \times 25}{(0.05)^{2}} \\ &= \frac{250}{0.0025} = 1 \times 10^{5}\ \text{N m}^{-1} \end{aligned} \]

The compression must be in metres (5 cm = 0.05 m). \(2 \times 10^{5}\) comes from dropping one of the factors of \(\frac{1}{2}\).

Answer: D

The work done is 8 J.

For a variable force, work is the integral of the force over the displacement.

\[ \begin{aligned} W &= \int_{0}^{2} 3x^{2}\,dx \\ &= \left[x^{3}\right]_{0}^{2} = 8\ \text{J} \end{aligned} \]

Multiplying the force at x = 2 m (12 N) by 2 m gives 24 J, and taking just F(2) = 12 gives the 12 J option; both treat the force as constant. 6 J is \(3x\) at 2 m, the derivative's mistake.

Answer: C

The power becomes 4P.

The pump gives kinetic energy to the water. In time t a mass \(m = \rho A v t\) leaves the hose, so the power is

\[ \begin{aligned} P &= \frac{\tfrac{1}{2}mv^{2}}{t} = \tfrac{1}{2}\rho A v^{3} \\ A &= \pi r^{2} \Rightarrow P \propto r^{2} \end{aligned} \]

With the speed constant, doubling the radius makes the area, and so the power, 4 times larger. 2P treats power as proportional to the radius, and 8P or 16P would need the speed to change too.

Answer: E

The block travels 25 m.

Kinetic friction gives a constant deceleration \(a = \mu_{k}g = 0.2 \times 10 = 2\ \text{ms}^{-2}\). With \(v^{2} = u^{2} - 2as\) and final speed 0:

\[ s = \frac{u^{2}}{2a} = \frac{100}{2 \times 2} = 25\ \text{m} \]

The mass cancels, so 2 kg is not needed. Forgetting the 2 in \(2as\) gives 50 m, which is not offered.

Answer: D

A non-conservative force is one for which the work depends on the path.

For a conservative force, such as gravity or a spring force, the work done between two points is the same whatever path is taken, and is zero around a closed loop, which is why a potential energy can be defined. Friction and air resistance are non-conservative: dragging a box along a longer route does more work against friction.

Work depending only on the end points describes a conservative force, not one that depends on the initial or final position alone. The work of a non-conservative force is not always zero, and mechanical energy is not conserved when such forces act; it is converted to heat.

Answer: A

The kinetic energy of the piece of mass M is \(\frac{3E}{4}\).

The body was at rest, so by conservation of momentum the two pieces move off with equal and opposite momenta \(p\). Kinetic energy is \(\frac{p^{2}}{2m}\), so for equal momenta it is inversely proportional to mass:

\[ \begin{aligned} \frac{K_{M}}{K_{3M}} &= \frac{3M}{M} = 3 \\ K_{M} &= \frac{3}{3 + 1}E = \frac{3E}{4} \end{aligned} \]

\(\frac{E}{4}\) is the share of the heavier piece. The lighter piece moves three times as fast and takes the larger share of the energy.

Answer: B

The kinetic energy is the lowest for the proton.

At the same speed, \(K = \frac{1}{2}mv^{2}\) is proportional to mass. The masses are in the ratio proton : deuteron : alpha particle = 1 : 2 : 4, so the kinetic energies are in the same ratio.

The proton therefore has the least kinetic energy and the alpha particle the most. That rules out 'same for all', 'same for proton and deuteron', 'highest for deuteron' and 'lowest for the alpha particle'.

Answer: B

The loss is 12.5 %.

The power available from the falling water is its loss of potential energy per second (taking g = 10 m/s²).

\[ \begin{aligned} P_{in} &= \frac{m}{t}gh = 24 \times 10 \times 100 = 24\ \text{kW} \\ \text{loss} &= \frac{24 - 21}{24} \times 100 = 12.5\% \end{aligned} \]

Dividing the 3 kW loss by the 21 kW output instead of the 24 kW input gives about 14.3%, which is not offered; 10% and 15% are rounded guesses.

Answer: C

The work done by gravity is Zero.

Gravity is a conservative force, so the work it does depends only on the start and end points. Going up a height h it does -mgh of work; coming back down it does +mgh. Over the round trip the displacement is zero and the total is zero.

It is not positive or negative overall, and although each half depends on the mass, the sum is zero for any mass.