Question

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. ​

Answer 1

$\large \sf \color{yellow}Your \: \: Answer : -$

Answer 2

Solution :-

• 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 - 4.8

Pi = - 0.8 kg m/s

Now ,

• 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 conserved

By 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