A softball of mass 0.220kg that is moving witha speed of8.5m/s collides head on and elastically with another ball initiallyat rest. Afterwards the incoming softball bounces backwardswith a speed of 3.7 m/s. Calculate: (a) the velocity ofthe target ball after the collision , and (b) the mass if te targetball.

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Alix Ortiz · Accepted Answer

The final speeds of the two balls are given by:  v=[M–mM+m]vi    wherevi is the initial speed of the smaller ball  V=[2mM+m]vi         note: I have used M for the bigger mass and m for the smaller mass,small v is the speed of the smaller ball and capital V is the speedof the larger ball.  Now using the first equation, you can find the largermass:  v=[M–mM+m]vior(M+m)v=(M–m)vi  Mv+mv=Mvi−mviorm(v+vi)=M(vi–v)  Finally:   M=m[v+vivi–v]=(0.220)[3.7+8.58.5–3.7]  So: M = 0.559kg  Then we can plug this back into the second equation at thetop to find V:  V=[2mM+m]vi=[2⋅0.2200.559+0.220]8.5=4.8ms  Hope this helps... please rate me!Thanks!       using conservation of momentum we get  0.22kg(8.5ms)+m2(0)=0.22kg(−3.7ms)+m2v2 .........1  for perfectelastic collision the coefficient of restitution = 1  therefore  v2−v1=−(u2−u1)  from this we get  a)       v2−(−3.7ms)=−(0−8.5ms)  from this we get v2     b)   substituting v2 in 1 we get    m2

A softball of mass 0.220kg that is moving witha speed of8.5m/s collides head on and elastically with another ball initiallyat rest. Afterwards the inc

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