Question

A block ofass m=4kg is moving with the velocity V=6m/s toward a target block of mass m=2kg,which is stationary (v=0).The objects collide head-on, and immediately, the speed of block MIS 8 times the speed of block M. What is the speed of block M?

Answer 1

To solve this problem, let's denote the speed of block M as v_M. We are given the following information:

Mass of block M (m_M) = 4 kg Velocity of block M (v_M) = ? Mass of block N (m_N) = 2 kg Initial velocity of block N (v_N) = 0 m/s Final velocity of block N (v_N') = 8 times the speed of block M (8v_M)

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:

(m_M * v_M) + (m_N * v_N) = (m_M * v_M') + (m_N * v_N')

Here, v_M' represents the final velocity of block M after the collision, and v_N' represents the final velocity of block N after the collision.

Let's plug in the given values:

(4 kg * 6 m/s) + (2 kg * 0 m/s) = (4 kg * v_M') + (2 kg * (8v_M))

24 kg * m/s = 4 kg * v_M' + 16 kg * v_M

Simplifying the equation:

24 = 4v_M' + 16v_M

Now, we know that the speed of block N (v_N') is 8 times the speed of block M:

v_N' = 8v_M

Substituting this into the equation:

24 = 4v_M' + 16(1/8)v_N'

Simplifying further:

24 = 4v_M' + 2v_N'

Since v_N' is 0 (as the block N is stationary after the collision), we can simplify the equation further:

24 = 4v_M'

Dividing both sides by 4:

6 = v_M'

Therefore, the speed of block M after the collision is 6 m/s.