Physics 9702 · AS & A Level · Equations of motion

Equations of motion — practice question

A steel ball is launched horizontally from the top of a table, as shown in Fig. 2.1. It leaves at a speed of $4.9\,\text{m s}^{-1}$. The ball lands on the ground a horizontal distance of $180\,\text{cm}$ from the table edge. Take air resistance as negligible.
(a(i))[1]

Calculate how long the ball takes to get to the ground.

(a(ii))[2]

Calculate the vertical velocity component of the ball when it reaches the ground.

(a(iii))[3]

Determine the magnitude and the angle to the horizontal of the velocity of the ball as it hits the ground.

(b)

The ball is launched by a compressed spring attached to a fixed block, as shown in Fig. 2.2. The ball is positioned on a frictionless track in front of the spring. It is then pulled back so that the spring has compression $x_0$. When the spring is released, the ball is projected horizontally as shown in Fig. 2.3.

(b(i))[3]

The ball is a uniform sphere of steel of diameter $0.016\,\text{m}$ and mass $0.017\,\text{kg}$. Calculate the density of the steel.

(b(ii))[2]

Every bit of the elastic potential energy in the spring becomes kinetic energy of the ball. The ball leaves the spring at a speed of $4.9\,\text{m s}^{-1}$. Show that the maximum elastic potential energy of the spring is $0.20\,\text{J}$.

(b(iii))[2]

Use Fig. 2.4 to determine the spring constant $k$ of the spring.

(b(iv))[2]

Use your answer in (b)(iii) together with the energy from (b)(ii) to find the compression $x_0$ of the spring.

(c(i))[1]

The steel ball is swapped for a polystyrene ball with the same diameter but a much smaller mass. The spring is compressed by $x_0$ and then released. Air resistance on this ball is not negligible after it leaves the spring. Explain why this ball leaves the spring with a greater speed than that of the steel ball.

(c(ii))[1]

Explain why this ball takes a longer time to reach the ground than the steel ball.

Worked solution & mark scheme

This 17-mark question has a full step-by-step worked solution and mark scheme. One marking point: Therefore, $t = 1.8 / 4.9 = 0.37\ \text{s}$.

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