Question

There are two marbles. The mass of 1st marble is 50g and the mass of 2nd marble is 80g. Before collision, the velocity of 1st marble is 8 m/s and the velocity of 2nd marble is 2m/s . Both the marble collider and after collision the (final) velocity of 1st marble becomes 4m/s Now calculate the final velocity of 2nd marble . Hint. Mi = 50g) M2=80) Before Collision Collidel After collision d 2 ms-1 (1=50) 2:8 g /​

Answer 1

m(12)+m(0)=mv1+mv2=0⇒v1+v2=12

−12−(0)v1−v2=1⇒v1−v2=−12

thus,v1=0v2=12,sovelocityoffirstaftercollisionis0

Answer 2

Answer:

4.5 m/s

Explanation:

Given-

mass of marble one (m1) = 50 g = 0.05kg

mass of marble two (m2) = 80 g = 0.08 kg

initial velocity of marble one(u1)= 8 m/s

initial velocity of marble two(u2)= 2 m/s

final velocity of marble one (v1) = 4 m/s

To find-

final velocity of marble two (v2)

We know that -

m1u1 + m2u2 = m1v1 + m2v2 (law of conservation of momentum)

putting the values -

0.05x8 + 0.08x2 = 0.05x4 + 0.08xv2

0.4+0.16=0.2+0.08xv2

0.56-0.2=0.08xv2

0.36=0.08xv2

0.36/0.08= v2

4.5= v2

hence final velocity of marble two is 4.5 m/s