You are playing a game where you throw a 200 gram bean bag horizontally onto a 1

Question

You are playing a game where you throw a 200 gram bean bag horizontally onto a 100 gram red ball; then, that red ball rolls 1 meter onto a at surface with a coecient of kinetic friction between the ball and the surface of 0.2. The red ball then makes a perfectly elastic collision with a 50 gram blue ball. The blue ball then rolls up on a frictionless 1-meter high ramp. To win, the blue ball must go as high as possible without falling o the ramp. At the same time, to win, the red ball must not touch the bean bag back. The bean bag hits the red ball and stay in place (its velocity is 0 after the collision). With what speed must you throw the bean bag to win?

Explanation / Answer

Let the initial velocity of bean bag be v,

initial velocity of red ball = 2v by momentum conservation,

kinetic energy of red ball just before collision = 0.5*0.1*(2v)^2 - 0.2*0.1*9.8*1 = 0.2v^2 - 0.196

velocity of red ball just before collision u = sqrt(4v^2 - 3.92)

velocity of red ball just after collision = u(0.100-0.050)/(0.100+0.050) = u/3

Now Final KE of red ball has to be zero so it does not collide with bean bag,

0.5*m*(u/3)^2 -0.196 = 0

u^2/9 = 0.196*2/0.1 = 3.92

[4v^2 - 3.92]/9 = 3.92

4v^2 - 3.92 = 3.92*9

4v^2 = 3.92*9+3.92 = 39.2

v^2 = 9.8

v = sqrt(9.8)

= 3.13 m/s answer

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