enter image description here

Why is (m)(v)=Impulse or as they put it here, Vs = I/Ms?

Shouldn't (m)(v) be equal to momentum, not I?

I don't understand why that is the solution. I was trying to solve for relative velocities, V2f - V1f, using momentum conservation but I failed hopelessly. I was thinking perhaps I should set momentum equal to zero so I have a center of mass frame, but I do not know how I should approach that. How and Why is this solution true?


Let $\mathbf F_s(t)$ and $\mathbf F_b(t)$ denote the forces exerted on the sattelite and booster during the explosion respectively. Let the explosion take place during the time interval $[t_1, t_2]$, then the impluses experienced by the satellite and booster during the explosion equal the changes in their momenta; \begin{align} \mathbf I_s &= \int_{t_1}^{t_2}dt\,\mathbf F_s(t) =\int_{t_1}^{t_2}dt\,\dot{\mathbf p}_s(t) = \mathbf p_s(t_2) - \mathbf p_s(t_1) = \Delta \mathbf p_s\\ \mathbf I_b &= \int_{t_1}^{t_2}dt\,\mathbf F_b(t) =\int_{t_1}^{t_2}dt\,\dot{\mathbf p}_b(t) = \mathbf p_b(t_2) - \mathbf p_b(t_1) = \Delta \mathbf p_b \end{align} If we assume that there are no external forces on the system, then momentum must be conserved, namely there should be no net change in momentum; \begin{align} \Delta \mathbf p_s + \Delta\mathbf p_b = 0 \end{align} so the impulses on the satellite and booster must be equal and opposite; \begin{align} \mathbf I_s = -\mathbf I_b \end{align} Of course, in reality this is an approximation because if there are other pieces that break off during the explosion, then strictly speaking, their momenta need to be accounted for as well. We assume these extra pieces have negligible comparable momenta. We also assume that the "impulse of the explosion" to which the problem refers is the magnitude of each of these impulses and is denoted by $I$.

Since there are only two objects, we can treat the final momenta of the system one-dimensionally (they will be equal and opposite), the impulse-momentum relationships above then tell us that \begin{align} m_s v_s = I, \qquad m_b v_b = -I \end{align} where we have taken $v_s(t_1) = 0$, $v_b(t_1) = 0$, $v_s(t_2) = v_s$, and $v_b(t_2) = v_b$. The rest is algebra.

  • $\begingroup$ You couldn't have given a better answer. $\endgroup$ Oct 31 '13 at 7:33

Firstly, remember that impulse= change in momentum= mv-mu

Suppose the satellite and the rocket before explosion was moving with velocity of u

Thus total momentum= 950*u+ 640*u

During explosion,impulse exerted on each other is the same (350Ns) but opposite in direction (Remember, Newton's third law states that the forces exerted on each other is same. Thus change in momentum caused by the force is same)

Two equations can be written by using the value of impulse:

For the rocket, I= -(640(v1)-640(u))






giving you relative velocity

I guess you got confused because the solution does not show how you do not need the initial velocity for your calculation


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.