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

Answer:

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

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 :

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

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