Question

If a ball of mass 0.4 kg moving with a velocity od 3 m/s collides elastically with another ball of mass 0.6 kg which is at rest find their velocities after collision

Answer 1

here , m1 = 0.4 kg
u1 = 3 m/s
m2 = 0.6 kg
u2 = 0 m/s
v1 and v2 => to be determined



using law of conservation of momentum
we have
(m1×u1)+(m2×u2) = (m1×v1)+(m2×v2)

=> (0.4×3) + (0.6×0) = v1(m1+m2)

as , when the bodies will collide their velocities will be equal in this case as ball is in rest , so both will move with Same velocity

=> 1.2+0 = v2(0.4+0.6)
=> 1.2 = v2(1)

=> 1.2 m/s = v2
=> 1.2 m/s = v1


hope this helps

Answer 2

According to law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
(0.4 × 3) + 0 = (0.4v1) + (0.6v2)
0.4v1 + 0.6v2 = 1.2
Multiply both sides by 10
4v1 + 6v2 = 12
Divide both sides by 2
2v1 + 3v2 = 6 …………[1]

Coefficient of restitution (e) is given as
e = (v2 - v1) / (u1 - u2)
For perfectly elastic collision e = 1
v2 - v1 = u1 - u2
v2 - v1 = 3 - 0
v2 - v1 = 3
Multiply both sides by 2
2v2 - 2v1 = 6 ……………[2]

Add equations [1] and [2]
2v1 - 2v1 + 3v2 + 2v2 = 6 + 6
5v2 = 12
v2 = 2.4 m/s

If v2 is 2.4 m/s then from equation [1] v1 = -0.6

After collision
Velocity of 0.4 kg ball = -0.6 m/s
Velocity of 0.6 kg ball = 2.4 m/s

After collision both moves in opposite direction.