Question

A 0.6 kg block moving with 6 m/s towards the right on a smooth horizontal plane collides with a 0.9 kg block moving with 2 m/s towards the left. If after the collision, the 0.9 kg block moves towards the right with 2.5 m/s, find the coefficient of restitution.

Answer 1

13/32

Answer 2

The problem involves a collision between two blocks. We will use the principle of conservation of linear momentum and the definition of the coefficient of restitution.

1. Define Variables and Given Information: Let $m_1$ and $m_2$ be the masses of the two blocks, and $u_1, u_2$ be their initial velocities, and $v_1, v_2$ be their final velocities. Given:

• Mass of the first block, $m_1 = 0.6$ kg
• Initial velocity of the first block, $u_1 = +6$ m/s (towards the right)
• Mass of the second block, $m_2 = 0.9$ kg
• Initial velocity of the second block, $u_2 = -2$ m/s (towards the left)
• Final velocity of the second block, $v_2 = +2.5$ m/s (towards the right) We need to find the final velocity of the first block, $v_1$, and the coefficient of restitution, $e$.

2. Apply Conservation of Linear Momentum: For a collision on a smooth horizontal plane, the total linear momentum of the system is conserved.

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Substitute the given values: $(0.6 \text{ kg})(+6 \text{ m/s}) + (0.9 \text{ kg})(-2 \text{ m/s}) = (0.6 \text{ kg})v_1 + (0.9 \text{ kg})(+2.5 \text{ m/s})$ $3.6 - 1.8 = 0.6v_1 + 2.25$ $1.8 = 0.6v_1 + 2.25$ $0.6v_1 = 1.8 - 2.25$ $0.6v_1 = -0.45$ $v_1 = \frac{-0.45}{0.6} = \frac{-45}{60} = -\frac{3}{4} = -0.75$ m/s

The negative sign indicates that the 0.6 kg block moves towards the left after the collision with a speed of 0.75 m/s.

3. Apply the Definition of Coefficient of Restitution ($e$): The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach.

$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$

Substitute the velocities:

$e = \frac{2.5 \text{ m/s} - (-0.75 \text{ m/s})}{6 \text{ m/s} - (-2 \text{ m/s})}$ $e = \frac{2.5 + 0.75}{6 + 2}$ $e = \frac{3.25}{8}$

To express $e$ as a fraction: $e = \frac{3.25}{8} = \frac{325}{800}$ Divide both numerator and denominator by 25: $e = \frac{325 \div 25}{800 \div 25} = \frac{13}{32}$

Alternatively, as a decimal: $e = 0.40625$

The coefficient of restitution is $\frac{13}{32}$ or $0.40625$.

Explanation of the solution:

1. Conservation of Momentum: The total momentum before collision equals the total momentum after collision. This allows us to find the unknown final velocity of the first block. $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ $(0.6)(6) + (0.9)(-2) = (0.6)v_1 + (0.9)(2.5)$ $1.8 = 0.6v_1 + 2.25 \implies v_1 = -0.75 \text{ m/s}$.
2. Coefficient of Restitution: This is the ratio of relative velocity of separation to relative velocity of approach. $e = \frac{v_2 - v_1}{u_1 - u_2}$ $e = \frac{2.5 - (-0.75)}{6 - (-2)} = \frac{3.25}{8} = \frac{13}{32}$.

Answer: The coefficient of restitution is $\frac{13}{32}$.