Question

A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball

Answer 1

Answer:u=-22 m/s

Step-by-step explanation:

Given

mass of puck
`m=0.06\ kg`

Average force
`f_(avg)=1.5* 10^3\ N`

time of contact
`t=1.2ms=1.2* 10^(-3)\ s`

puck leaves with a velocity of
`v=8\ m/s`

We know impulse is
`F_(avg)\Delta t`
`=\text{change in momentum}`

therefore

`1.5* 10^3* (1.2* 10^(-3))=P_f-P_i`

`P_i=0.06* 8-1.8`

`P_i=0.48-1.8=-1.32\ kg-m/s`

Final momentum
`P_f=m* v_f`

`P_f=0.06* 8`

`P_f=0.48\ kg-m/s`

Impulse on the ball
`=F_(avg)\Delta t`

Impulse
`=1.5* 10^3* 1.2* 10^(-3)=1.8\ N-s`

Initial velocity is given by

`u=(P_i)/(m)=(-1.32)/(0.06)`

`u=-22\ m/s`

i.e. initially ball is moving towards -x-axis