Question

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

• A. 0.5
• B. 0.8
• C. 0.25
• D. 0.4

Answer 1

The Correct Option is C

To solve the problem of determining the coefficient of restitution, let us analyze the collision between the two blocks.

Given:

• Mass of the moving block, m_1 = m
• Mass of the stationary block, m_2 = 4m
• Initial velocity of the moving block, u_1 = v
• Initial velocity of the stationary block, u_2 = 0
• Velocity of the moving block after collision, v_1 = 0

Since the moving block comes to rest after the collision, we need to find the velocity of the second block after collision, v_2.

The law of conservation of momentum states that the total momentum before collision equals the total momentum after collision:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Substitute the given values:

mv + 4m \cdot 0 = m \cdot 0 + 4mv_2

Simplify to find v_2:

v_2 = \frac{v}{4}

The coefficient of restitution e is defined as:

e = \frac{v_2 - v_1}{u_1 - u_2}

Substitute the known values:

e = \frac{\left(\frac{v}{4} - 0\right)}{v - 0} = \frac{v/4}{v}

Simplify further:

e = \frac{1}{4} = 0.25

Thus, the correct answer is 0.25.