# 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

Relevance

- Anonymous8 months agoFavorite Answer
m*ΔV = F*Δt

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

- MyRankLv 67 months 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

Source(s): https://myrank.co.in/ - Anonymous8 months ago
change in momentum = impulse

change in momentum = mass * change in speed

impulse = force * time

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