Question

A 15 kg block is sliding along a frictionless surface and strikes a 10 kg ball at rest. What is the collision of the blocks after the collision if it is an inelastic collision?

Answer 1

In an inelastic collision between a 15 kg block and a 10 kg ball, the two objects stick together and move as one after the collision. The resulting mass is 25 kg, and the velocity of the combined mass after the collision is 1.6 m/s.

Step-by-step explanation:

In an inelastic collision, the two objects stick together and move as one after the collision.

In this case, the 15 kg block and the 10 kg ball collide and stick together. The resulting mass is the sum of the two masses, which is 15 kg + 10 kg = 25 kg.

Since it is an inelastic collision, the velocity of the combined mass after the collision is due to the conservation of momentum. The total momentum before the collision is (15 kg * 0 m/s) + (10 kg * 4 m/s) = 40 kg*m/s.

Since the total mass after the collision is 25 kg, the velocity of the combined mass after the collision is 40 kg*m/s ÷ 25 kg = 1.6 m/s.

Answer 2

v = 0.6 v₁

Step-by-step explanation:

This is an exercise in collisions, let's start by defining a system formed by the two bodies, so that the forces during the collisions have been internal and the momentum is preserved.

Instant starts. Before the crash

p₀ = M v₁ + m 0

Final moment. After the crash

`p_(f)` = (M + m) v

how momentum is conserved

p₀ = p_{f}

M v₁ = (M + m) v

v =
`(M)/(M+m) v_(1)`

let's calculate

v =
`(15)/(15+10) v_(1)`

v = 0.6 v₁