Question

A block (mass 5 kg ) sits at rest. A second block (mass 10 kg ) collides with the first block, initially moving with a speed of 12 m/s just before the collision. After the collision, the two blocks stick together, and they slide for 4 s across a rough surface before coming to a stop. What was the magnitude of the average force of friction acting on the blocks after the collision?

Answer 1

Answer: The magnitude of the average force of friction acting on the blocks after the collision is 30 N.

Explanation:

According to the conservation of momentum,

    $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{f}$

Putting the given values into the above formula as follows.

     $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{f}$

     $0 + 10 \times 12 = (5 + 10)v_{f}$

         $v_{f} = \frac{12 \times 10}{15}$ m/s

We assume that acceleration due to friction is $-a_{f}$. Also,

           v = u + at

Here,  v = 0 m/s

          u = 8 m/s

          t = 4 sec

Then, value of acceleration will be as follows.

           v = u + at

      $0 = 8 + a \times 4$

       a = 2 $m/s^{2}$

Now, we know that expression for frictional force is as follows.      

    $F_{f} = (m_{1} + m_{2}) \times a_{f}$

               = $15 \times 2$

              = 30 N

Thus, we can conclude that the magnitude of the average force of friction acting on the blocks after the collision is 30 N.