Two bumper cars on a frictionless surface collide elastically.The first bumper car is moving to the right with a speed of 22.0 m/s and rear-ends a bumper car also moving to the right with a speed of 5.00 m/s.What is the speed of the first bumper car after the collision (in m/s)? The mass of the first bumper car is 192 kg and the mass of the second bumper car is 144 kg.Assume that the collision takes place in one dimension.
A)3.9643
B)4.6382
C)5.4267
D)6.3492
E)7.4286

Question

Quizplus · Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. The formula for conservation of momentum is $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$, where $m_1$ and $m_2$ are the masses of the cars, $v_{1i}$ and $v_{2i}$ are the initial velocities, and $v_{1f}$ and $v_{2f}$ are the final velocities. For kinetic energy conservation, the equation is $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$.Given: $m_1 = 192\,kg$, $m_2 = 144\,kg$, $v_{1i} = 22.0\,m/s$, $v_{2i} = 5.00\,m/s$.Using the conservation of momentum:$$192 \times 22.0 + 144 \times 5.00 = 192v_{1f} + 144v_{2f}$$And knowing that in an elastic collision, the relative speed of approach before the collision is equal to the relative speed of separation after the collision, we can solve for $v_{1f}$ and $v_{2f}$ using these two conservation laws. However, without going through the detailed algebra, the correct answer is obtained by solving these equations, which results in $v_{1f} = 7.4286\,m/s$. This is a complex problem that typically requires solving a system of equations, but the key takeaway is that the final speed of the first car can be found by applying conservation laws, leading to the correct answer.

Two Bumper Cars on a Frictionless Surface Collide Elastically