Question

A small, 150 g cart is moving at 1.40 m/s on a frictionless track when it collides with a larger, 1.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision?

Answer 1

Approximately
`0.345\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Multiply the velocity of an object by its mass to find the momentum. In other words, when an object of mass
`m` travels at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

Under the assumptions (friction is negligible, track is horizontal), the combined momentum of the two objects would be "conserved" (stay the same) during the collision.

Ensure that mass is measured in the standard unit (kilograms):

`150\; {\rm g} = 0.150\; {\rm kg}`.

Before the collision, combined momentum of the two objects (in
`{\rm kg\cdot m\cdot s^(-1)}`) was:

`\begin{aligned}p(\text{before}) &= (0.150)\, (1.40) + (1.00)\, (0.890) = 0.210\end{aligned}`.

Note that since
`0.890\; {\rm m\cdot s^(-1)}` is the recoil speed of the
`0.150\; {\rm kg}` object, this object would be moving backwards. The actual velocity of this object should bd
`(-0.890)\; {\rm m\cdot s^(-1)}`.

Let
`v` denote the velocity of the
`1.00\; {\rm kg}` object after the collision (in
`{\rm m\cdot s^(-1)}`.) Combined momentum after the collision (also in
`{\rm kg\cdot m\cdot s^(-1)}`) would be:

`\begin{aligned}p(\text{after}) &= (0.150)\, (-0.890) + (1.00)\, v \end{aligned}`.

Since the total momentum (in
`{\rm kg\cdot m\cdot s^(-1)}`) is conserved:

`p(\text{after}) = p(\text{before}) = 0.210`.

`\begin{aligned} (0.150)\, (-0.890) + (1.00)\, v = 0.210\end{aligned}`.

Rearrange and solve for the unknown velocity
`v`:

`v \approx 0.345\; {\rm m\cdot s^(-1)}`.