A ball of mass 0.5 kg moving with a velocity of 2 m/s strikes a wall normally and bounces back with the same s

Question

A ball of mass 0.5 kg moving with a velocity of 2 m/s strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:

Accepted Answer

1. **Identify Given Values:**
   - Mass ($m$) = 0.5 kg
   - Initial velocity ($v_i$) = $+2$ m/s (towards the wall)
   - Final velocity ($v_f$) = $-2$ m/s (bounces back with same speed, opposite direction)
   - Time of contact ($Δt$) = 1 millisecond = $1 	imes 10^{-3}$ s

2. **Calculate Change in Momentum (Impulse):**
   Impulse is defined as the change in momentum ($Δp = p_f - p_i$).
   $$ \Delta p = m(v_f - v_i) $$
   $$ \Delta p = 0.5 	imes (-2 - 2) = 0.5 	imes (-4) = -2 	ext{ kg m/s} $$
   The magnitude of the impulse is $2 	ext{ kg m/s}$ [Source 66, 68].

3. **Apply Newton's Second Law:**
   The average force ($F_{avg}$) is the rate of change of momentum.
   $$ F_{avg} = \frac{|\Delta p|}{\Delta t} $$
   $$ F_{avg} = \frac{2 	ext{ kg m/s}}{1 	imes 10^{-3} 	ext{ s}} = 2000 	ext{ N} $$
   [Source 66]

Step-by-Step Solution

Practice All PHYSICS Questions