Question

Cart 1 with mass 4.44 kg collides with a stationary cart (cart 2) of mass 2.81 kg. Before the collision cart 1 has a velocity of 4.94 m/s to the right. After the collision, cart 1 has a velocity 1.11 m/s to the right and cart 2 has a velocity 6.05 m/s to the right.

A) What is the total momentum of the two cart system before the collision?
B) What is the total kinetic energy of the two cart system before the collision?
C) What is the total momentum of the two cart system after the collision?
D) What is the total kinetic energy of the two cart system after the collision?
E) Is this collision elastic or inelastic? Be wary of round-off errors here.

Answer 1

Answer:

a) $p = 21.934\,\frac{kg\cdot m}{s}$, b) $E=54.176\,J$, c) $p = 21.929\,\frac{kg\cdot m}{s}$, d) $E=54.161\,J$, e) A perfectly elastic collision (e = 1).

Explanation:

a) The total momentum of the two cart system before the collision is:

$p = (4.44\,kg)\cdot (4.94\,\frac{m}{s} )+(2.81\,kg)\cdot (0\,\frac{m}{s} )$

$p = 21.934\,\frac{kg\cdot m}{s}$

b) The total kinetic energy of the two cart system before the collision is:

$E = \frac{1}{2}\cdot (4.44\,kg)\cdot (4.94\,\frac{m}{s} )^{2} + \frac{1}{2}\cdot (2.81\,kg)\cdot (0\,\frac{m}{s} )^{2}$

$E=54.176\,J$

c) The total momentum of the two cart system after the collision is:

$p = (4.44\,kg)\cdot (1.11\,\frac{m}{s} )+(2.81\,kg)\cdot (6.05\,\frac{m}{s} )$

$p = 21.929\,\frac{kg\cdot m}{s}$

d) The total kinetic energy of the two cart system after the collision is:

$E = \frac{1}{2}\cdot (4.44\,kg)\cdot (1.11\,\frac{m}{s} )^{2} + \frac{1}{2}\cdot (2.81\,kg)\cdot (6.05\,\frac{m}{s} )^{2}$

$E=54.161\,J$

e) The coefficient of restitution, which is useful to distinguish elastic collisions from inelastic ones is given by following ratio:

$e = \frac{E_{f}}{E_{o}}$

$e = \frac{54.161\,J}{54.176\,J}$

$e = 1$

Which is a perfectly elastic collision.