Question

Two carts mounted on an air track are moving toward one another. Cart 1 has a speed of 0.8 m/s and a mass of 0.47 kg. Cart 2 has a mass of 0.61 kg.
(a) If the total momentum of the system is to be zero, what is the initial speed of Cart 2? m/s
(b) Does it follow that the kinetic energy of the system is also zero since the momentum of the system is zero?
(c) Determine the system's kinetic energy in order to substantiate your answer to part (b).

Answer 1

Answer:

a) 0.62 m/s

b) No

c) 0.27 joules

Explanation:

a) Total linear momentum (P) is the sum of the individual momenta of Cart 1 and Cart 2:

$\overrightarrow{p}=\overrightarrow{p_{1}}+\overrightarrow{p_{2}}$

if it's zero

$\overrightarrow{p_{1}}+\overrightarrow{p_{2}}=0$

Momentum is the product of mass and velocity so:

$m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}} = 0$

because the direction of the Carts is opposite to each other, we should use a negative sing for one of the velocities (no matter which)

$m_{1}v_{1}+m_{2}(-v_{2}) =0$

$m_{1}v_{1}-m_{2}v_{2}=0$

solving for $v_{2}$

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{m_{1}v_{1}}{m_{2}}=\frac{(0.47)(0.8)}{0.61}$

$v_{2}=0.62\frac{m}{s}$

b) No, kinetic energy and linear momentum are different things, one of the most important differences is that kinetic energy is a scalar quantity and linear momentum is a vector quantity.

c)System kinetic energy is the sum of the individual kinetic energies

$K=K_{1}+K{2}=\frac{m_{1}v_{1}^{2}}{2}+\frac{m_{2}v_{2}^{2}}{2}$

$K= \frac{(0.47)(0.8)^{2}}{2}+\frac{(0.61)(0.62)^{2}}{2}=0.27J$

It's different of zero!