# A 531 g ball strikes a wall at 15.6 m/s and rebounds at 10.4 m/s. ?

The ball is in contact

with the wall for 0.032 s.

What is the magnitude of the average force

acting on the ball during the collision?

Answer in units of N.

### 3 Answers

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- Anonymous2 months agoFavorite Answer
m*ΔV = F*Δt

F = m*ΔV/Δt = 0.531*(15.6+10.4)*1000/32 = 431 N

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- MyRankLv 63 weeks ago
Given,

Mass (m) = 531grams = 0.531kg

Initial speed (u) = 15.6m/sec

Final speed (v) = 10.4m/sec

Time (t) = 0.032sec

Magnitude of average force (F) = ?

We know that:-

Impulse (I) = change in momentum (∆p)

Force (F) x time (t) = mass x change in velocity

Force (F) = 0.531kg x (15.6 + 10.4)m/sec / 0.032sec = 431N

∴ Force (F) = 431N

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- Anonymous2 months ago
change in momentum = impulse

change in momentum = mass * change in speed

impulse = force * time

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