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A mass A of 0.8 kg moving to the right with a speed of 5 m/s collides head- on with a ball B of 1.2 kg moving in the opposite direction with a speed of 4 m/s. After the collision, A is moving to the left with a speed of 4 m/s. Find the velocity of B after the collision and the coefficient of restitution. |
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Answer» n :-Mass of A(Ma) = 0.8 kg Speed of A(Va) = 5 m /s Mass of B (Mb)= 1.2 kg Speed of B(VB) = - 4 m/s ( opposite direction)Initial momentum of the system (Pi)Pi = Pa +Pb Pi = Ma x Va + Ma x Vb Pi = 0.8 x 5 + 1.2 x -4 Pi = 4 - 4.8Pi = - 0.8 kg m/sNow ,Speed of mass A = - 4 m/s Final momentum of the system (Pf)Pf = Pa' +Pb 'Pf = Ma x Va '+ Ma x Vb'Pf = 0.8 x -4 + 1.2 Vb' Pf = -3.2 + 1.2 Vb 'As no external force is acting on the system the total momentum of the system is conservedBy applying LAW of conservation of momentum ,Pi = Pf - 0.8 = - 3.2 + 1.2Vb'1.2Vb' = 2.4 Vb '= 2 m/s The ball will move with a velocity of 2 m/s towards the RIGHT Coefficient of restitution e = Vb' - Va' / Va - Vb e = 2 -(-4) / 5-(-4)e = 6 / 10 e = 0.6 The coefficient of restitution is 0.6 |
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