Question

1. a 2.00 kg rock is dropped from the top of a 30.0 m high building. calculate the balls momentum at the time that it strikes the ground.
2. a 925 kg car moving at a velocity of 19.0 m/s right collides with a stationary truck of unknown mass. the two vehicles lock together and move off at a velocity of 6.50 m/s. what is the mass of the truck?
3. a 46 gram tennis ball is launched from a 1.35 - kg homemade cannon. if the cannon recoils with a speed of 2.1 m/s, determine the muzzle speed of the tennis ball.

Explanation:

Step1: <Find the velocity of the rock when it strikes the ground>

We use the kinematic equation \(v^{2}=v_{0}^{2}+2a\Delta y\). Since the rock is dropped (\(v_{0} = 0\)), \(a = g=9.8\ m/s^{2}\) and \(\Delta y=30.0\ m\).

$$v=\sqrt{2g\Delta y}=\sqrt{2\times9.8\times30}$$

$$v=\sqrt{588}\approx24.25\ m/s$$

Step2: <Calculate the momentum>

The formula for momentum is \(p = mv\). Given \(m = 2.00\ kg\) and \(v\approx24.25\ m/s\)

$$p=(2.00\ kg)\times(24.25\ m/s)=48.5\ kg\cdot m/s$$

Step1: <Apply the law of conservation of momentum>

The law of conservation of momentum states \(m_{1}v_{1}+m_{2}v_{2}=(m_{1} + m_{2})v_{f}\). Here, \(m_{1}=925\ kg\), \(v_{1}=19.0\ m/s\), \(v_{2} = 0\) (truck is stationary), and \(v_{f}=6.50\ m/s\)

$$m_{1}v_{1}=(m_{1}+m_{2})v_{f}$$

$$925\times19=(925 + m_{2})\times6.5$$

$$17575=6.5\times925+6.5m_{2}$$

$$17575 = 6012.5+6.5m_{2}$$

$$6.5m_{2}=17575 - 6012.5$$

$$6.5m_{2}=11562.5$$

$$m_{2}=\frac{11562.5}{6.5}=1779\ kg$$

Step1: <Apply the law of conservation of momentum>

The law of conservation of momentum states \(m_{1}v_{1}+m_{2}v_{2}=0\) (initial momentum is zero). Let \(m_{1}=1.35\ kg\), \(v_{1}=- 2.1\ m/s\) (recoil velocity) and \(m_{2}=0.046\ kg\)

$$m_{1}v_{1}=-m_{2}v_{2}$$

$$v_{2}=-\frac{m_{1}v_{1}}{m_{2}}$$

$$v_{2}=-\frac{1.35\times(-2.1)}{0.046}$$

$$v_{2}=\frac{2.835}{0.046}\approx61.6\ m/s$$

Answer:

\(48.5\ kg\cdot m/s\)