Answer to Question #237821 in Mechanics | Relativity for ala

Question

A mass m1=6kg is moving on a smooth horizontal frictionless surface at 2m/s west when collide with an equal mass m at rest. Determine (a) the total momentum of the system after interaction (b) the momentum of m2 after interaction (c) the velocity of m2 after interaction (d) assuming that m1 and m2 are in contact for 0.05sec during interaction, what is the average net force applied to m2 by m1 during that time?

Answer 1

(a) We can calculate the initial momentum and this will be equal to the momentum at the end of the collision:

$\\ p_{i, T}=m_1v_{1,i}+m_2v_{2,i}=m(v_{1,i}+v_{2,i}) \\ p_{i, T} =(6\,kg)[(-2\,m/s)+(0\,m/s)] \\ p_{i, T} =-12\frac{kg\cdot m}{s}=p_{f,T} \text{ (conservation of momentum)} \\ p_{f, T}=m_1v_{1,f}+m_2v_{2,f}=m(v_{1,f}+v_{2,f}) \\ -12\frac{kg\cdot m}{s} =(6\,kg)[v_{1,f}+v_{2,f}] \\ v_{1,f}+v_{2,f}= -2\frac{m}{s}$

Then, since the sum of the two velocities has to be equal to -2 m/s, this would mean that the kinetic energy is also conserved:

$\frac{1}{2}mv^2_{1,i}=\frac{1}{2}mv^2_{1,f}+\frac{1}{2}mv^2_{2,f} \\ \implies v^2_{1,i}=v^2_{1,f}+v^2_{2,f}$

We can use the equation that relates the final velocities and elevate the whole expression to the square exponent and then we can have another equation that relates v_1,f and v_2,f:

$\\ [v_{1,f}+v_{2,f}]^2= [-2\frac{m}{s}]^2 \\ \implies v^2_{1,f}+v^2_{2,f}+2v_{1,f}v_{2,f}=4\frac{m}{s^2}$

Once we substitute on the equation that comes from the conservation of kinetic energy:

$\\ \implies v^2_{1,i}+2v_{1,f}v_{2,f}=4\frac{m^2}{s^2} \\ \implies (-2\,\frac{m}{s})^2+2v_{1,f}v_{2,f}= 4\frac{m^2}{s^2} \\ 4\frac{m^2}{s^2}=2v_{1,f}v_{2,f}+4\frac{m}{s^2} \to 2v_{1,f}v_{2,f}=0\frac{m^2}{s^2}$

This last solution implies that one of the blocks has to stop moving while the other takes all the kinetic energy and starts traveling. In conclusion, we would suggest that m₁ stopped moving while m₂ started to move and then:

$\\ v_{1,f}+v_{2,f}= -2\frac{m}{s} \\ \text{since } v_{1,f}=0\frac{m}{s}; v_{2,f}=-2\frac{m}{s} \\ \text{With this we find the momentum:} \\ p_{2,f}=mv_{2,f}=(6\,kg)(-2\frac{ m}{s}) \\ \implies p_{2,f}=-12\frac{kg\cdot m}{s}$

Now, we only need to find the average net force applied if we know that for mass m₂:

$F=\dfrac{\Delta p}{\Delta t}=\dfrac{p_f-p_i}{\Delta t}=\dfrac{-12\frac{kg\cdot m}{s}-0\frac{kg\cdot m}{s}}{0.05\,s} \\ \text{ } \\ \implies F= 240\,N$

In conclusion, we found that (a) the total momentum of the system after the interaction is -12 kgm/s, (b) the momentum of m₂ after the interaction is -12 kgm/s, (c) the velocity of m₂ after the interaction is -2.0 m/s, and (d) the average net force applied to m₂ by m₁ is about 240 N.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.