**Problem 7:** A 4 kg traveling with a speed of 2 m/s
strike a rigid wall and rebounds elastically. If the ball is in contact with
the wall for 0.05 sec, what is

- the momentum imparted to the wall
- the average force exerted on the wall

**Solution**

**Given: **Mass of the ball, m = 4 kg, v_{i} = 2 m/s, v_{f} = -2 m/s, Δt = 0.05 sec

**Find: **Momentum imparted (passed on) to the wall
and the average force on the wall.

Note

- The collision is elastic, therefore, net change in the momentum will be zero.
- If the velocity toward the wall is taken positive, the rebounded velocity will be negative.

Now initial momentum = p_{i} = mv_{i}
= 4 × 2 = 8 kg ms^{-1}

Final momentum =
p_{f} = mv_{f} = 4 × (-2) = – 8 kg ms^{-1}

Therefore, change
in momentum of the ball = Δp = p_{f} – p_{i} = – 8 – 8 = – 16
kg ms^{-1}

This change in the momentum of the ball is passed on to the wall so that the change in the momentum of the system is zero.

**To find the
average force on the wall:**

From Newton’s 2^{nd}
law of motion, F = ma = m (v/t). Therefore, momentum expressed as Newton’s 2^{nd}
law is, F× t = mv = p. Therefore, change in momentum = Δp = mv_{f} – mv_{i}
= m (v_{f} – v_{i}). Hence,

F × Δt = m × Δv. Put the values,

F × 0.05 = 2 × [-8 – 8)] = -2 × 16 = -16

OR F = 16/0.05 = -320 N

Hence, the average force exerted on the wall is 320 N.

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