Question

A 4.6 kg bowling ball sliding to the right at 7.43 m/s has an elastic head-on collision with another 4.6 kg bowling ball initially at rest. The first ball stops after the collision. Find the velocity of the second ball after the collision. Answer in units of m/s.

Answer 1

V₂ = 7.43 m/s

Step-by-step explanation:

To find the velocity of the second ball after the collision, we will use the following formula:

`(M_1*U_1) + (M_2*U_2) = (M_1*V_1) + (M_2*V_2)`

where M₁ = mass of ball 1 = 4.6 kg

M₂ = mass of ball 2 = 4.6 kg

U₁ = velocity of ball 1 before collision = 7.43 m/s

U₂ = velocity of ball 2 before collision = 0 m/s

V₁ = velocity of ball 1 after collision = 0 m/s

V₂ = velocity of ball 1 after collision = ?

So substituting the given values in the above formula to get:

(4.6 × 7.43) + (4.6 × 0) = (4.6 × 0) + (4.6 × V₂)

34.178 = (4.6)V₂

V₂ = 7.43 m/s

Answer 2

Ans) 7.43m/sec

Exp: Here we have to apply conservation of momentum formula

i.e. m1u1 + m2u2 = m1v1 + m2v2

where m1 &m2 are the mass of first and second particle respectively and u1 and u2 are the initial velocity and v1 and v2 are the final velocities of respective masses respectively

Putting all the values,

4.6×7.43 + 4.6×0 = 4.6×0 + 4.6×v2

i.e. v2 = 7.43 m/sec.

Hope it helps!!!