A football of mass 0.42kg is passed with a velocity of `25 ms^-1` due south. A defending player lunges at the ball and deflects it so that the new velocity is `20 ms^-1` `36.9^@` west of south . If the player is in contact with the ball for 0.05s , what is the magnitude of the average force he exerts

Options

A. 63N
B. 126N
C. 75N
D. 105N

Answer

B

Explanation

We choose south and west as positive Final velocity has components `16 ms^-1` south and `12 ms^-1` west
Initial momentum `=0.42 times 0.25=10.5 kg ms^-1` south
Final momentum `=(0.42 times 16)` south and `(0.42 times 12)` west
`[Delta mv]_(south)=0.42 (16-25) =-3.78 kg ms^-1`
`[Delta mv]_(west)=0.42 times 12=5.04 kg ms^-1`
`|Delta mv|=[(5.04)^2+(3.78)^2]^(1//2)=6.3 kg ms^-1`
Impulse `=Delta mv= 6.3 Ns`
Impulse `=Ft=0.05 vec F=6.3 N s`
`vec F=6.3 Ns//0.05s=126N`