ps07sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2010 Problem Set 7: Conservation of Momentum and Collisions Solutions Problem 1 A slender uniform rod of length L and mass M rests along the y-axis on a frictionless, horizontal table. A particle of equal mass M is moving along the x-axis at a speed V 0 . At t = 0 the particle strikes the end of the rod and sticks to it. Find the position ! R cm ( t ) and velocity ! V cm ( t ) of the center of mass of the system as a function of time. Solution: The center of mass of the system of rod and particle is given by the expression ! R cm ( t ) = m 1 ! r 1 ( t ) + m 2 ! r 2 ( t ) m 1 + m 2 . where ! r 1 ( t ) is the position of the center of mass of the rod and ! r 2 ( t ) is the position of the particle. The center of mass of the rod for t ! 0 is given by the expression ! r 1 ( t ) = ( L / 2) ˆ j ; t ! 0 . For time t ! 0 , the position of the particle is ! r 2 ( t ) = V 0 t ˆ i ; t ! 0 .
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So the center of mass of the system at time t ! 0 is ! R cm ( t ! 0) = MV 0 t ˆ i + M ( L / 2) ˆ j 2 M = ( V 0 t / 2) ˆ i + ( L / 4) ˆ j . The velocity of the center of mass is given by ! V cm ( t ) = m 1 ! v 1 ( t ) + m 2 ! v 2 ( t ) m 1 + m 2 . The rod is at rest for t ! 0 and so ! v 1 ( t ! 0) = ! 0 . For time t ! 0 , the particle is moving in the positive x-direction and the velocity is given by ! v 2 ( t ! 0) = V 0 ˆ i . So the velocity of the center of mass of the system at time t ! 0 is given by ! V cm ( t ) = MV 0 ˆ i 2 M = ( V 0 / 2) ˆ i ; t ! 0 . For t > 0 , because there are no external forces acting on the system, the velocity of the center of mass of the system is constant and the position of the center of mass of the system moves according to ! R cm ( t ) = ! R cm ( t = 0) + ! V cm t = ( L / 4) ˆ j + ( V 0 / 2) t ˆ i ; t > 0 ! V cm ( t ) = 0 ˆ i 2 M = ( V 0 / 2) ˆ i ; t > 0 .
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