Answer to Question #174500 in Physics for bailey

Question #174500

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.


1
Expert's answer
2021-03-23T11:10:39-0400

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


p=m1v1p = m_1v_1

is equal to the total momentum after the collision:


p=m1v1+m2v2p' = m_1v_1' + m_2v_2'

where m1=40kg,m2=29kgm_1 = 40kg, m_2 = 29kg are the masses of the objects, v1=25m/sv_1 = 25m/s is the speed of the first object before the collision, and v1,v2v_1', v_2' are their speeds after the collsion. Thus, obtain:


m1v1=m1v1+m2v2m_1v_1 = m_1v_1' + m_2v_2'


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


m1v122=m1v122+m2v222\dfrac{m_1v_1^2}{2} = \dfrac{m_1v_1'^2}{2} +\dfrac{m_2v_2'^2}{2}

Expressing v1v_1' from the first equation and substituting it into the last one, after doing some algebra obtain:


v22(m22m1+m2)2m2v1v2=0v22(m2m1+1)2v1v2=0v2(v2(m2m1+1)2v1)=0v_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 v2=0v_2' = 0, which is not possible. Another one is:

v2(m2m1+1)2v1=0v2=2v1m2m1+1=2m1v1m1+m2v_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:


v2=2402540+2928.99 m/sv_2'=\dfrac{2\cdot 40\cdot 25}{40+29} \approx 28.99 \space m/s

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