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…

1.

Explanation:

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

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

Answer:

$6.16\ m/s$

2.

Explanation:

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.

Answer:

$- 6.67\ m/s$

3.

Explanation:

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:

$2\ km/h$

4.

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the system is $p_i=(m_{Joe + Bo}+m_{car1})v_1$, where $m_{Joe + Bo}=200.0\ kg$, $m_{car1}=100.0\ kg$ and $v_1 = 10.0\ m/s$. The final momentum of the system is $p_f=(m_{Joe + Bo})v_2+m_{car2}v_3$, where $m_{car2}=m_{Melinda}+m_{car1}$, $m_{Melinda}=25.0\ kg$, $v_2 = 4.12\ m/s$. According to the law of conservation of momentum $p_i=p_f$.

Step2: Expand and solve for $v_3$

$(200 + 100)\times10=200\times4.12+(25 + 100)v_3$. $3000=824 + 125v_3$. $125v_3=3000 - 824=2176$. $v_3=\frac{2176}{125}=17.408\ m/s$

Answer:

$17.4\ m/s$

5.

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the system is $p_i=m_{sports - car}v_{sports - car}+m_{pick - up}v_{pick - up}$, where $m_{sports - car}=800\ kg$, $v_{sports - car}=13.0\ m/s$, $m_{pick - up}=1200\ kg$ and $v_{pick - up}=25.0\ m/s$. After the collision, the combined mass is $m_f=m_{sports - car}+m_{pick - up}$. According to the law of conservation of momentum $p_i=p_f$, or $m_{sports - car}v_{sports - car}+m_{pick - up}v_{pick - up}=(m_{sports - car}+m_{pick - up})v_f$.

Step2: Solve for the final velocity

$v_f=\frac{m_{sports - car}v_{sports - car}+m_{pick - up}v_{pick - up}}{m_{sports - car}+m_{pick - up}}=\frac{800\times13+1200\times25}{800 + 1200}=\frac{10400+30000}{2000}=\frac{40400}{2000}=20.2\ m/s$

Answer:

$20.2\ m/s$