Question

Two blocks are sliding along a frictionless track. Block A (mass 4.03 kg) is moving to the right at 3.00 m/s. Block B (mass 4.84 kg) is moving to the left at 3.60 m/s. Assume the system to be both Block A and Block B. What is the total momentum of the system before the collision?

Answer 1

The total momentum of the system before the collision is 5.334 kg-m/s towards left.

Step-by-step explanation:

Given that,

Mass of the block A,
`m_A=4.03\ kg`

Speed of block A,
`v_A=3\ m/s`

Mass of the block B,
`m_B=4.8\ kg`

Mass of block B,
`u_B=-3.6\ m/s`

Let p is the total momentum of the system before the collision. It is given by :

`p=m_Av_A+m_Bv_B\\\\p=4.03* 3+4.84* (-3.6)\\\\p=-5.334\ kg-m/s`

So, the total momentum of the system before the collision is 5.334 kg-m/s towards left. Hence, this is the required solution.