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

0.40625

Answer 2

The problem asks to find the coefficient of restitution for a collision between two blocks. We will use the principles of conservation of linear momentum and the definition of the coefficient of restitution.

1. Define the variables and directions: Let the direction to the right be positive (+). 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)

After collision: Final velocity of the second block, $v_2 = +2.5$ m/s (towards the right) Let the final velocity of the first block be $v_1$.

2. Apply the principle of conservation of linear momentum: The total linear momentum before the collision is equal to the total linear momentum after the collision. $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Substitute the known values into the equation: $(0.6 \text{ kg})(+6 \text{ m/s}) + (0.9 \text{ kg})(-2 \text{ m/s}) = (0.6 \text{ kg})v_1 + (0.9 \text{ kg})(+2.5 \text{ m/s})$ $3.6 - 1.8 = 0.6v_1 + 2.25$ $1.8 = 0.6v_1 + 2.25$ $0.6v_1 = 1.8 - 2.25$ $0.6v_1 = -0.45$ $v_1 = \frac{-0.45}{0.6} = -0.75 \text{ m/s}$

The negative sign indicates that the first block moves towards the left after the collision.

3. Apply the formula for the 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 calculated and given: $u_1 = +6$ m/s $u_2 = -2$ m/s $v_1 = -0.75$ m/s $v_2 = +2.5$ m/s

$e = \frac{2.5 - (-0.75)}{6 - (-2)}$ $e = \frac{2.5 + 0.75}{6 + 2}$ $e = \frac{3.25}{8}$ $e = 0.40625$

The coefficient of restitution is approximately 0.406.

The final answer is $\boxed{0.40625}$.

Explanation of the solution:

1. Conservation of Momentum: The total momentum of the system before collision ($m_1 u_1 + m_2 u_2$) is equated to the total momentum after collision ($m_1 v_1 + m_2 v_2$). This equation is used to find the unknown final velocity ($v_1$) of the first block.
2. Coefficient of Restitution: The coefficient of restitution ($e$) is calculated using the formula $e = \frac{v_2 - v_1}{u_1 - u_2}$, which represents the ratio of the relative velocity of separation to the relative velocity of approach. All velocities must be substituted with their correct signs based on the chosen positive direction.