Two astronauts X and Y float in deep space, far from external forces. Initially, they are at rest relative to one another. X throws a tool towards Y. Consider the two astronauts and the tool as one system. Which one of the following statements is correct?
The centre of mass remains fixed.
The two astronauts and the tool form an isolated system since they are far from all external forces. Before the throw, the net momentum of the system is zero. When X throws the tool towards Y, the forces involved are purely internal to the system (X on the tool, and later the tool on Y). Internal forces cannot change the total momentum of a system, so the total momentum stays zero throughout. Since the centre of mass moves with velocity \(v_{cm} = \dfrac{P_{total}}{M_{total}}\), and \(P_{total}=0\) remains constant, the centre of mass does not move at all — it stays fixed in space. Hence option (d) is correct.
The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its
A heavy particle of rest mass M while moving along the positive z - direction, decays into two identical light particles with rest mass m (where M > 2m). The maximum value of the momentum that any one of the lighter particles can have in a direction perpendicular to the z-direction, is
A metal ball with the momentum mv strikes a wall and bounces back. The change in the ball's momentum is ideally
A mass of 4 kg moving at 3 m/s collides with a mass of 6 kg moving at 2 m/s in the opposite direction and they stick together. The combined mass has velocity of:
In an orbital motion, the angular momentum vector is: