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Ball 1 is released from rest. It collides with block 2 in an elastic healon collision sending up the ramp. The level surface
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Answer #1

velocity of bob just before hitting box

u^2 = 2 gh = 2* 9.8 * 1

u = 4.427 m/s

velocity of box just after elastic collision

v = 2 m u / ( m + M) = 2* 10* 4.427 / ( 10 + 2)

v = 7.379 m/s

using conservation of energy

mg h + u mg d = 0.5 m v^2

9.8* d sin 40 + 0.6* 9.8* d = 0.5 * 7.379^2

d = 2.235 m

=======

Comment before rate in case any doubt, will reply for sure.. goodluck

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