question 6 a vehicle with a mass of 1,628 kg that is travelling at 6 m/s collides with a stationary vehicle…

question 6 a vehicle with a mass of 1,628 kg that is travelling at 6 m/s collides with a stationary vehicle that has a mass of 3,096 kg. assuming the vehicles have an inelastic collision, what will be the final velocities of the two vehicles? 1 point question 7 a vehicle with a mass of 1,629 kg that is travelling at 6.7 m/s collides with a stationary vehicle that has a mass of 3,365 kg. assuming the vehicles have an elastic collision and that the first vehicle comes to a stop, what will be the final velocities of the second vehicle? 1 point 3 of 10 questions remaining

question 6 a vehicle with a mass of 1,628 kg that is travelling at 6 m/s collides with a stationary vehicle that has a mass of 3,096 kg. assuming the vehicles have an inelastic collision, what will be the final velocities of the two vehicles? 1 point question 7 a vehicle with a mass of 1,629 kg that is travelling at 6.7 m/s collides with a stationary vehicle that has a mass of 3,365 kg. assuming the vehicles have an elastic collision and that the first vehicle comes to a stop, what will be the final velocities of the second vehicle? 1 point 3 of 10 questions remaining

Answer

Explanation:

Step1: Apply conservation of momentum for in - elastic collision

For an in - elastic collision, the formula is $m_1u_1+m_2u_2=(m_1 + m_2)v$. Here, $m_1 = 1628$ kg, $u_1=6$ m/s, $m_2 = 3096$ kg and $u_2 = 0$ m/s. Substitute the values into the formula: $1628\times6+3096\times0=(1628 + 3096)v$.

Step2: Solve for the final velocity $v$

First, calculate the left - hand side: $1628\times6=9768$ kg·m/s. The right - hand side is $(1628 + 3096)v=4724v$. Set them equal: $9768 = 4724v$. Then $v=\frac{9768}{4724}\approx2.068$ m/s.

Answer:

The final velocity of the two - vehicle system is approximately 2.068 m/s.

Explanation:

Step1: Apply conservation of momentum for elastic collision

For an elastic collision, $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ and $u_1 - u_2=v_2 - v_1$. Here, $m_1 = 1629$ kg, $u_1 = 6.7$ m/s, $m_2 = 3365$ kg, $u_2 = 0$ m/s and $v_1 = 0$ m/s. From $u_1 - u_2=v_2 - v_1$, we have $6.7-0=v_2 - 0$, so $v_2 = 6.7$ m/s is wrong. We use the conservation of momentum formula $m_1u_1+m_2u_2=m_1v_1+m_2v_2$. Substituting the values: $1629\times6.7+3365\times0=1629\times0+3365v_2$.

Step2: Solve for $v_2$

$1629\times6.7 = 3365v_2$. $v_2=\frac{1629\times6.7}{3365}=\frac{10914.3}{3365}\approx3.24$ m/s

Answer:

The final velocity of the second vehicle is approximately 3.24 m/s.