Question

A 140 g ball with speed $\mathbf{7}\mathbf{.}\mathbf{8}\frac{\mathbf{m}}{\mathbf{s}}$ strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. The collision lasts 3.80 ms .What are the magnitudes of the (a) impulse and (b) average force on the wall from the ball during the elastic collision?

Short answer

1. Magnitude of impulse,$J\mathrm{is}2.18\frac{\mathrm{kg}.\mathrm{m}}{\mathrm{s}}$.
2. Magnitude of the average force on the wall from the ball, F is 575 N .

Step-by-step solution

Understanding the given information

1. Mass of the ball, m = 140 g.
2. Speed of the ball,$v_i=7.8\frac{m}{s}$.
3. Time of collision, t = 3.80 ms .

Concept and formula used in the given question

Use the concept of impulse. Using the equation of impulse and find the impulse and then using the same equation of time involved, then find the average force. The equations are given below.

$$\begin{gathered} J=\mathrm{m}∆\mathrm{V} \\ Ft=\mathrm{m}∆\mathrm{V} \end{gathered}$$

(a) Calculate the magnitude of impulse

Magnitude of impulse:

Ball rebounds back with same velocity, we can write,

$$\begin{gathered} J=\left(0.14\right)(-v-(v) \\ =0.14\times 2v \\ =0.14\times 2\times 7.8 \\ =-2.18\frac{\mathrm{kg}.\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Magnitude of impulse is $J=2.18\frac{\mathrm{kg}.\mathrm{m}}{\mathrm{s}}$

(b) Calculate the magnitude of impulse average force on the wall from the ball during the elastic collision

Magnitude of the average force on the wall from the ball.

Consider the formulas $J=m∆V\mathrm{and}Ft=m∆V$.

Use value of J here and solve as:

$$\begin{gathered} F(3.80\times {10}^{-3})=m∆V \\ F=\frac{2.18}{3.80\times {10}^{-3}} \\ F=574.7\mathrm{N} \\ F\approx 575\mathrm{N} \end{gathered}$$