A ball of mass 0.2 kg is thrown normally against a wall with a speed of 15m/s and rebound from there with a velocity 20m/s. The impulse of the force exerted by the ball on the wall is
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Answered by
33
Impulse=change in momentum
P2-p1
=mv-(-mu)
=0.2*20+0.2*15
=4+3
=7
P2-p1
=mv-(-mu)
=0.2*20+0.2*15
=4+3
=7
chiragverma:
thanks sir
is this formula
MV-(-mu )
Answered by
1
Explanation:
impulse =change in momentum
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