Answer to Question #168948 in Physics for Froilan

Question

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.60 m/s and that of the trailing car is 6.00 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds?

v_{leading_bumper_car} =

v_{trailing_bumper_car} =

Answer 1

Let the mass of the leading bumper car be $m_1$ and the mass of the trailing bumper car be $m_2$. Let the initial speed of the leading bumper car be $u_1$ and the initial speed of the trailing bumper car be $u_2$. From the law of conservation of momentum, we have:

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2. (1)$$

Since collision is elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2. (2)$$

Let’s rearrange equations (1) and (2):

$$m_1(u_1-v_1)=m_2(v_2-u_2), (3)$$ $$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)$$

Let’s divide equation (4) by equation (3):

$$\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},$$ $$u_1+v_1=u_2+v_2. (5)$$

Let's express $v_2$ from the equation (5) in terms of $u_1$, $u_2$ and $v_1$:

$$v_2=u_1-u_2+v_1. (6)$$

Let’s substitute equation (6) into equation (3). After simplification, we get:

$$(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1.$$

From this equation we can find the final speed of the leading bumper car, $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(400\ kg-400\ kg)}{(400\ kg+400\ kg)}\cdot5.6\ \dfrac{m}{s}+\dfrac{2\cdot400\ kg}{(400\ kg+400\ kg)}\cdot6\ \dfrac{m}{s},$$ $$v_1=6\ \dfrac{m}{s}.$$

The sign plus means that the leading bumper car moves to the right after collision.

Substituting $v_1$into the equation (6) we can find the final speed of the trailing bumper car, $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\cdot400\ kg}{(400\ kg+400\ kg)}\cdot5.6\ \dfrac{m}{s}+\dfrac{(400\ kg-400\ kg)}{(400\ kg+400\ kg)}\cdot6\ \dfrac{m}{s},$$ $$v_2=5.6\ \dfrac{m}{s}.$$

The sign plus means that the trailing bumper car moves to the right after collision.