Aracquet ball with mass m = 0.241 kg is moving toward the wall at v = 13.7 m/s and at an angle of θ = 34° 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 = 0.063 s. 1) what is the magnitude of the initial momentum of the racquet ball? 2) what is the magnitude of the change in momentum of the racquet ball? 3) what is the magnitude of the average force the wall exerts on the racquet ball?

Explanation:

1 )  Initial momentum in x ( horizontal direction ) = mv cos34

= .241 x 13.7 cos 34

= 2.73  kg m/s

=  Initial momentum in Y-xis ( vertical direction ) = mv sin 34  kg m/s

= 1.85 kg m/s

2 )   Final momentum in x ( horizontal direction ) = - mv cos34

= -  .241 x 13.7 cos 34

= -  2.73  kg m/s

=  Final  momentum in Y-xis ( vertical direction ) = mv sin 34  kg m/s

= 1.85 kg m/s

change in momentum in x direction

= 2.73 - (-2.73 )

= 5.46 kg m/s

Change in momentum in Y direction

= 1.85 - 1.85 = 0

3 ) If average force is F

impulse = change in momentum

F x 0.063 = 5.46

F = 86.67 N.

pecialized keyboards

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm