Question

1. A ball of mass m1=0.39kg moving east (+x direction) with a speed of 5m/s collided head on with ball m2 with mass 0.27kg ball which is at rest. If the collision is perfectly Elastic, what will be the speed of the ball m1 after the collision?
2. A sled with a mass of 24.8kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 6.9−m rope to a post set in the ice. Once given a push, the sled revolves uniformly in a circle around the post. If the sled makes five complete revolutions every minute, find the force F exerted on it by the rope.
3. A car of mass 831-kg, as shown in the figure below, is pushed along a floor with a force of 478.4-N for a distance of 8.9-m. Calculate the work done.

Answer 1

1) From the law of conservation of momentum, we have:

$$m_1v_{1i}=m_1v_{1f}+m_2v_{2f}.$$

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

$$\dfrac{1}{2}m_1v_{1i}^2=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2.$$

This formula gives us an additional relationship between velocities. Therefore, with the help of these two formulas we can find the velocity of the first ball ball after collision:

$$v_{1f}=\dfrac{m_1-m_2}{m_1+m_2}v_{1i},$$ $$v_{1f}=\dfrac{0.39\ kg-0.27\ kg}{0.39\ kg+0.27\ kg}\times5\ \dfrac{m}{s}=0.91\ \dfrac{m}{s}.$$

2) We can find the force F exerted on the sled by the rope as follows:

$$F=ma=\dfrac{mv^2}{r}=m\omega^2r,$$ $$F=24.8\ kg\times(5\ \dfrac{rev}{min}\times\dfrac{1\ min}{60\ s}\times\dfrac{2\pi\ rad}{1\ rev})^2\times6.9\ m,$$ $$F=47\ N.$$

3) By the definition of the work done, we have:

$$W=Fd=478.4\ N\times8.9\ m=4258\ J.$$