Question

A ball having a mass of 0.5 kg is moving towards the east with a speed of 8.0 ms-1.
After being hit by a bat it changes its direction and starts moving towards the north with
a speed of 6.0 ms−1. If the time of impact is 0.1 s, calculate the impulse and average
force acting on the ball.

Answer 1

Answer on Question#74093 – Physics – Mechanics

A ball having a mass of $0.5\,\mathrm{kg}$ is moving towards the east with a speed of $8.0\,\mathrm{ms}^{-1}$. After being hit by a bat it changes its direction and starts moving towards the north with a speed of $6.0\,\mathrm{ms}^{-1}$. If the time of impact is $0.1\,\mathrm{s}$, calculate the impulse and average force acting on the ball.

**Solution.**

We use the concept of linear momentum

$$\vec{p} = m \vec{v}$$

where $m$ is a mass of the ball, $\vec{v}$ is a velocity of the ball. (for our case)

By definition impulse is the change in momentum. Hence we can write

$$\Delta \vec{p} = \overrightarrow{p_f} - \overrightarrow{p_i} = m \Delta \vec{v} = m \left( \overrightarrow{v_f} - \overrightarrow{v_i} \right)$$

where $\overrightarrow{p_f}$ is final momentum, $\overrightarrow{p_i}$ is initial momentum, $\overrightarrow{v_f}$ is final velocity, $\overrightarrow{v_i}$ is initial velocity.

According to the condition of problem

$\overrightarrow{v_i} = 8.0\,\frac{m}{s}$ directed towards the east; $\overrightarrow{v_f} = 6.0\,\frac{m}{s}$ directed towards the north; $m = 0.5\,\mathrm{kg}$ is a mass of the ball.

Draw a speed sketch

From a right triangle we find the magnitude of the vector $\Delta \vec{v}$

$$|\Delta \vec{v}| = \sqrt{|\overrightarrow{v_i}|^2 + |\overrightarrow{v_f}|^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\,\frac{\mathrm{m}}{\mathrm{s}}.$$

Therefore impulse

$$\Delta \vec{p} = m \Delta \vec{v} = 0.5\,\mathrm{kg} \cdot 10\,\frac{\mathrm{m}}{\mathrm{s}} = 5\,\mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}.$$

On the other hand, we can calculate the impulse as

$$\Delta \vec{p} = \vec{F} t$$

where $\vec{F}$ is average force acting on the ball, $t = 0.1\,\mathrm{s}$ is time of impact. Hence

$$\vec{F} = \frac{\Delta \vec{p}}{t} = \frac{5\,\mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}}{0.1\,\mathrm{s}} = 50\,\mathrm{N}$$

Answer. $\Delta \vec{p} = 5\,\mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}$; $\vec{F} = 50\,\mathrm{N}$

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