Question

A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed vi= 8.4 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 What is V, the speed of the second block after the collision?

Answer 1

Answer:

$v_{2'}=8.1\:\mathrm{m/s}$

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:

$\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot8.4^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot 4.4\cdot {v_{2'}}^2,\\\\{v_{2'}}=\sqrt{64.31},\\\\{v_{2'}}\approx\fbox{ 8.1\:\mathrm{m/s} }$..