Question

Dynamics cart 1 has a mass of 0.84 kg and is
initially moving at 4.2 m/s
. Cart 1 undergoes
an elastic head-on collision with dynamics cart 2.

The mass of cart 2 is 0.48 kg, and cart 2 is initially
moving at 2.4 m/s
. The collision is cushioned
by a spring with spring constant 8.0 3 103
N/m. T/I
(a) Calculate the final velocity of each cart after
they completely separate.
(b) Determine the compression of the spring
during the collision at the moment when cart 1
is moving at 3.0 m/s
.
(c) Determine the maximum compression of
the spring.

Answer 1

(a) Since the collision is elastic, you have conservation of momentum and conservation of kinetic energy.

\[m _{1}u _{1}+ m _{2}u _{2} = m _{1}v _{1} + m _{2}v _{2}\]

\[\frac{ 1 }{ 2 }m _{1}u _{1}^{2} + \frac{ 1 }{ 2 }m _{2}u _{2}^{2} = \frac{ 1 }{ 2}m _{1}v _{1}^{2}+\frac{ 1 }{ 2}m _{2}v _{2}^{2} \]

You can calculate the final velocities of the carts

(b)

Given \[v _{1} = 4 m/s\]

Calculate \[v _{2}\]

with the equation

\[m _{1}u _{1}+ m _{2}u _{2} = m _{1}v _{1} + m _{2}v _{2}\]

Then, calculate x the compression of the spring with the equation

\[\frac{ 1 }{ 2 }m _{1}u _{1}^{2} + \frac{ 1 }{ 2 }m _{2}u _{2}^{2} = \frac{ 1 }{ 2}m _{1}v _{1}^{2}+\frac{ 1 }{ 2}m _{2}v _{2}^{2} + \frac{ 1 }{ 2}kx ^{2}\]

(c)

At the point of maximum compression of the spring, the 2 carts will have the same velocity,

Use the momentum equation to calculate this velocity

and the equation to calculate the max compression

\[\frac{ 1 }{ 2 }m _{1}u _{1}^{2} + \frac{ 1 }{ 2 }m _{2}u _{2}^{2} = \frac{ 1 }{ 2}m _{1}v _{f}^{2}+\frac{ 1 }{ 2}m _{2}v _{f}^{2} + \frac{ 1 }{ 2}kx _{m} ^{2}\]