Answer to Question #174500 in Physics for bailey

Question

Object A has a mass of 40 kg and is traveling at 25 m/s [W] when it strikes a stationary object B with a mass of 29 kg in a perfectly elastic collision. Determine the velocity of object B after the collision.

Answer 1

In a perfectly elastic collision the momentum conservation law holds. Thus, the total momentum before the collision:

$$p = m_1v_1$$

is equal to the total momentum after the collision:

$$p' = m_1v_1' + m_2v_2'$$

where $m_1 = 40kg, m_2 = 29kg$ are the masses of the objects, $v_1 = 25m/s$ is the speed of the first object before the collision, and $v_1', v_2'$ are their speeds after the collsion. Thus, obtain:

$$m_1v_1 = m_1v_1' + m_2v_2'$$

Futhermore, the energy conservation law also holds for an elastic collision:

$$\dfrac{m_1v_1^2}{2} = \dfrac{m_1v_1'^2}{2} +\dfrac{m_2v_2'^2}{2}$$

Expressing $v_1'$ from the first equation and substituting it into the last one, after doing some algebra obtain:

$$v_2'^2\left( \dfrac{m_2^2}{m_1} + m_2 \right)-2m_2v_1v_2'=0\\ v_2'^2\left( \dfrac{m_2}{m_1} +1 \right)-2v_1v_2'=0\\ v_2' \left( v_2'\left( \dfrac{m_2}{m_1} +1 \right) - 2v_1 \right) = 0$$

One solution is $v_2' = 0$, which is not possible. Another one is:

$$v_2'\left( \dfrac{m_2}{m_1} +1 \right) - 2v_1 = 0\\ v_2' = \dfrac{2v_1}{\dfrac{m_2}{m_1} +1} = \dfrac{2m_1v_1}{m_1+m_2}$$

Substituting the numbers, obtain:

$$v_2'=\dfrac{2\cdot 40\cdot 25}{40+29} \approx 28.99 \space m/s$$