Answer to Question #281131 in Physics for xdx

Question

1. A 0.45-caliber bullet with m = 0.133 kg hurled from a gun’s muzzle at 950 m/s.

2. A 0.126-lb tennis ball has been shot up from rest with a racket. Suppose the ball reaches a height of 5.5 m. Calculate the net impulse of the force

 

3. On average, a standard 57.0-g tennis ball is said to be in contact with the tennis racket for 30 milliseconds. Suppose that a serve is done with a speed of 262.8 kph. What amount of force was exerted on the ball during the serve?

 

4. A block with mass equal to 12 000 g slid across a frictionless surface at 37.0 m/s and hit another block at rest. After the impact, they bounced off of each other with respective velocities of 42.502 mph and 8.948 mph. Determine the mass of the second block.

5. Two balls of masses 10.0 kg and 20.0 kg collided such that the 10.0-kg ball moved with a velocity of 15.66 mph. If the second ball was at rest prior to the collision, what is its final velocity?

 6. Infer the importance of the Conservation of Momentum.

 

Answer 1

1)

$$p=mv=0.133\ kg\times950\ \dfrac{m}{s}=126.35\ \dfrac{kg\times m}{s}.$$

2)

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv_i^2,$$ $$v_i=\sqrt{2gh}=\sqrt{2\times9.8\ \dfrac{m}{s^2}\times5.5\ m}=10.38\ \dfrac{m}{s},$$ $$J=m\Delta v=m(v_f-v_i),$$ $$J=0.057\ kg\times(0-10.38\ \dfrac{m}{s})=-0.6\ \dfrac{kg\times m}{s}.$$

3)

$$F\Delta t=\Delta p,$$ $$F=\dfrac{\Delta p}{\Delta t}=\dfrac{mv}{\Delta t},$$ $$F=\dfrac{0.057\ kg\times73\ \dfrac{m}{s}}{30\times10^{-3}\ s}=138.7\ N.$$

4) Let's apply the law of conservation of momentum:

$$m_1v_{1i}=-m_1v_{1f}+m_2v_{2f},$$ $$m_2=\dfrac{m_1(v_{1i}+v_{1f})}{v_{2f}},$$ $$m_2=\dfrac{12\ kg\times(37\ \dfrac{m}{s}+19\ \dfrac{m}{s})}{4\ \dfrac{m}{s}}=168\ kg.$$

5) 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 addititional relationship between velocities. Therefore, with

the help of these two formulas we can find the velocity of the second ball after the collision:

$$v_{2f}=\dfrac{2m_1v_{1i}}{m_1+m_2}=\dfrac{2\times10\ kg\times6.7\ \dfrac{m}{s}}{10\ kg + 20\ kg}=4.47\ \dfrac{m}{s}.$$

6) The law of conservation of momentum states that the total momentum of an isolated system (if no external force acting on it) remains constant. Hence, the importance of the law of conservation of momentum is that we can find the final momentum of the body, or its final velocity (if we know its initial velocity before the collision).