Question

What is the velocities of car b after the collision

Answer 1

Given,

The mass of the bumper car A is

`M_A=281\text{ kg}`

The mass of the bumper car B is

`M_B=209\text{ kg}`

The velocity of the bumper car A is

`V_A=2.82\text{ m/s}`

The velocity of the bumper car B is

`V_B=1.72\text{ m/s}`

To find: The velocity of car B after the collision.

Let v₁ and v₂ denote the velocity of car A and car B, respectively, after the collision. It sounds like both cars are initially moving in the same direction (since both have positive initial velocity).

Since momentum is conserved,

`\begin{gathered} (281\text{ kg\rparen}*(2.82\text{ m/s\rparen+\lparen209 kg\rparen}*(1.72\text{ m/s\rparen=281 kg}* v_1+209\text{ kg}* v_2 \\ 1151.9\text{ kg.m/s= 281v}_1+209v_2.............(1) \end{gathered}`

The kinetic energy is also conserved.

`\begin{gathered} (1)/(2)*281*(2.82)^2+(1)/(2)*209*(1.72)^2=(1)/(2)*281* v_1^2+(1)/(2)*209* v_2^2 \\ 2852.93\text{ }(m^2)/(s^2)=281v_1^2+209v_2^2..........(2) \end{gathered}`

Solve the first equation for v₁ :

`v_1=(1151.9-209v_2)/(281)`

Substitute v₁ into the second equation and solve for v₂.

`\begin{gathered} 2852.93\text{ }(m^2)/(s^2)=281[(1,151.9-209v_(2))/(281)]^2+209v_2^2 \\ v_2^2=2.96\text{ }(m^2)/(s^2),\text{ 8.89 }(m^2)/(s^2) \\ v_2=1.72\text{ m/s, 2.98 m/s} \end{gathered}`

Where we ignore the first solution since it corresponds to the initial condition.

Thus, the velocity of car B after the collision is

`v_2=2.98\text{ m/s}`