A 655g ball strikes a wall at 14.9 m/s and rebounds at 13.4 m/s. The ball is in contact with the wall for 0.034 sec.
What is the magnitiude of the force acted on the ball during the collision? Answer in units of N.
2006-11-17 10:16:22 · 2 answers · asked by invu 1 in Science & Mathematics ➔ Physics
I'll assume the ball's velocity is normal to the wall. Then momentum change = impulse = (14.9+13.4)*.655 N-s. Divide this by .034 s to find force in N.
2006-11-17 10:46:50 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
(V2 - V1) / t = a
V1 = (+) 14.9
V2 = (-) 13.4
t = 0.034
a = acceleration = (-13.4 - (+)14.9) / 0.034 = (-13.4 - 14.9) / 0.034 = -832 m/s^2
Force = Mass x Acceleration = 0.655 Kg x (-832) = - 545 N
Force that acted on the ball is 545 Newtons in the opposite direction
2006-11-17 18:49:06 · answer #2 · answered by naike_10021980 2 · 1⤊ 0⤋