Question

A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s The block of mass M travels to the right at a speed V of 6.7 m/s what is M

Answer 1

Answer:

$m_2=6.3\:\mathrm{kg}$

Explanation:

In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:

$\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2$

Since the second block was initially at rest, $\frac{1}{2}m_2{v_2}^2=0$.

Plugging in all given values, we have:

$\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{ 6.3\:\mathrm{kg} }$.