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

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

Step-by-step explanation:

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

`(1)/(2)m_1{v_1}^2+(1)/(2)m_2{v_2}^2=(1)/(2)m_1{v_(1')}^2+(1)/(2)m_2{v_(2')}^2`

Since the second block was initially at rest,
`(1)/(2)m_2{v_2}^2=0`.

Plugging in all given values, we have:

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