A racquet ball with mass m = 0.239 kg is moving toward the wall at v = 13.1 m/s

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

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