Question

1. a 1250 kg car is stopped at a traffic light. a 3550 kg truck moving at 8.33 m/s hits the car from behind. if bumpers lock, how fast will the two vehicles move?
2. the muzzle velocity of a 50.0 g shell leaving a 3.00 kg rifle is 400. m/s. what is the recoil velocity of the rifle?
3. imagine that you are hovering next to a space shuttle and your buddy of equal mass who is moving a 4 km/h with respect to the ship bumps into you. if he holds onto you, how fast do you both move with respect to the ship?
4. joe and his brother bo have a combined mass of 200.0 kg and are zooming along in a 100.0 kg amusement park bumper car at 10.0 m/s. they bump into melindas car, which is sitting still. melinda has a mass of 25.0 kg. after the collision, the twins continue ahead with a speed of 4.12 m/s. how fast is melindas car bumped across the floor?
5. if an 800. kg sports car slows to 13.0 m/s to check out an accident scene and the 1200. kg pick - up truck behind him continues traveling at 25.0 m/s, with what velocity will the two move if they lock bumpers after a rear - end collision?

Explanation:

1.

Step1: Apply conservation of momentum

The initial momentum of the system is the momentum of the truck only since the car is at rest. The formula for momentum is $p = mv$. The initial momentum $p_i=m_{truck}v_{truck}$, where $m_{truck}=3550\ kg$ and $v_{truck}=8.33\ m/s$. After the collision, the two - vehicles stick together, so $m_f=m_{truck}+m_{car}$, with $m_{car} = 1250\ kg$. According to the law of conservation of momentum $p_i = p_f$, or $m_{truck}v_{truck}=(m_{truck}+m_{car})v_f$.

Step2: Solve for the final velocity

$v_f=\frac{m_{truck}v_{truck}}{m_{truck}+m_{car}}=\frac{3550\times8.33}{3550 + 1250}=\frac{29571.5}{4800}\approx6.16\ m/s$

Step1: Apply conservation of momentum

The initial momentum of the rifle - shell system is zero (as the system is at rest initially). According to the law of conservation of momentum $p_i=p_f$. Let the mass of the shell be $m_{shell}=50.0\ g=0.05\ kg$, its velocity $v_{shell}=400\ m/s$, and the mass of the rifle be $m_{rifle}=3.00\ kg$. The final momentum of the system is $m_{shell}v_{shell}+m_{rifle}v_{rifle}=0$.

Step2: Solve for the recoil velocity of the rifle

$v_{rifle}=-\frac{m_{shell}v_{shell}}{m_{rifle}}=-\frac{0.05\times400}{3.00}=-\frac{20}{3}\approx - 6.67\ m/s$. The negative sign indicates that the rifle moves in the opposite direction of the shell.

Step1: Apply conservation of momentum

Let your mass be $m$ and your buddy's mass be $m$ (equal mass). Your initial velocity $v_1 = 0\ km/h$ and your buddy's initial velocity $v_2=4\ km/h$. After the collision, the combined mass is $m + m=2m$. According to the law of conservation of momentum $m_1v_1+m_2v_2=(m_1 + m_2)v_f$. Substituting the values, we get $m\times0+m\times4=(m + m)v_f$.

Step2: Solve for the final velocity

$4m = 2mv_f$, so $v_f = 2\ km/h$

Answer:

$6.16\ m/s$

2.