Question

a 6kg trolley moving at 2m/s hits a stationary 4kg trolley. they stick together. What is their common velocity after the collision?

Answer 1

Answer on Question 60077, Physics – Mechanics | Relativity

Question:

A $6\,kg$ trolley moving at $2\,ms^{-1}$ hits a stationary $4\,kg$ trolley. They stick together. What is their common velocity after the collision?

Solution:

We can find the common velocity after the collision from the Law of Conservation of Momentum:

$$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{c},$$

here, $m_{1}, m_{2}$ are the masses of the first and second trolleys, respectively; $v_{1}, v_{2}$ are the initial velocities of the first and second trolleys before the collision, respectively; $v_{c}$ is their common velocity after the collision.

Since, the second trolley is at rest ($v_{2} = 0\,ms^{-1}$), we get:

$$m_{1}v_{1} = (m_{1} + m_{2})v_{c}.$$

From this formula we can find the common velocity of two trolleys after the collision:

$$v_{c} = \frac{m_{1}v_{1}}{m_{1} + m_{2}} = \frac{6\,kg \cdot 2\,ms^{-1}}{6\,kg + 4\,kg} = \frac{12\,kg\,ms^{-1}}{10\,kg} = 1.2\,ms^{-1}.$$

Answer:

$$v_{c} = 1.2\,ms^{-1}.$$