Question

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 24 m/s. The masses of the two objects are 3.8 and 8.4 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.

Answer 1

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$