A racquet ball with mass m = 0.239 kg is moving toward the wall at v = 13.1 m/s
ID: 2140916 • Letter: A
Question
A racquet ball with mass m = 0.239 kg is moving toward the wall at v = 13.1 m/s and at an angle of ? = 32
A racquet ball with mass m = 0.239 kg is moving toward the wall at v = 13.1 m/s and at an angle of ? = 32degree with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.06 s. What is the magnitude of the initial momentum of the racquet ball? kg-m/s What is the magnitude of the change in momentum of the racquet ball?Explanation / Answer
1) P1 = m*v1 = 3.1309 kg.m/s
2) P1x = m*v1*cos(32) = 2.655 kg.m/s
P2x = -m*v2*cos(32) = -2.655 kg.m/s
P1y = m*v1*sin(32) = 1.659 kg.m/s
P2y = m*v2*sin(32) = 1.659 kg.m/s
chnage in momentum = P2x - P1x = -5.31 kg.m/s
3)
Fav = (P2x-P1x)/time = -88.5 N
- sign indictes froce is in -x direction
4)
change in momntum = m*(Vf-Vi) = 0.239*(-8.5-13.1) = -5.1624 kg.m/s
Fav = (P2-P1)/time
==> time = (P2-P1)/Fav = -5.1624/(-88.5) = 0.05833 s
6)
chenge in KE = 0.5*m*(vf^2-Vi^2) = -11.87352 J