1as shown in figure?? [7]
A ball of mass M1=1kg falling vertically with a velocity v0=2m/s;strikes a wedge of mass M=2kg kept on a smooth,horizontal surface as shown in figure.The coefficient of restitution between the ball and the wedgel is e=1/2; Find the velocity of the wedge and the ball immediately after collision.
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5 Answers
33The coefficient of restitution between the ball and the ball is e=1/2
[3]
33This is a sort of standard question...
u got at max 3 variables...
vx of wedge and velocity of ball along normal direction...
and also there are 2 eqns..
1. momentum of system is conserved in horizontal direction..
2. coefficient of restitution...
I will post a complete soln...
33
velocities shown with red are vel after collision and blue earlier the collision...
initial momentum along horizontal direction of the system is zero
therefore...
Mvx-m((2sinθ)cosθ))-m(vsinθ)=0
now velocity of separation = e(velocity of approach).. of the point of contact..
therefore..
v+vxsinθ=e(2cosθ)
put the values of θ,m,M and e... (M is mass of wedge and m is of ball..)
putting the values..
v=vx=1/√3
so velocity of block 1i+1/√3 j where i is down the incline and j is perpendicularly upward from the incline...
there can be mistake in putting values... [3]