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.
- 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
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) = 431NSource(s): https://myrank.co.in/
- Anonymous8 months ago
change in momentum = impulse
change in momentum = mass * change in speed
impulse = force * time