One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 680 km above Earth's surface collides with a pellet having mass 3.2 g. (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a 3.2 g bullet from a modern army rifle with a muzzle speed of 730 m/s?

An object, with mass 88 kg and speed 18 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?

A block of mass mm = 16 kgkg has a speed VV and is behind a block of mass MM = 65 kgkg that has a speed of 0.5 m/sm/s. The surface is frictionless. The blocks collide and couple. After the collision, the blocks have a common speed of 0.9 m/sm/s. In the figure, the loss of kinetic energy of the blocks due to the collision is closest to: 7.7 J 25 J 26 J 9.3 J 76 J

We know that the mass of the bagged lunch is m = 1.40 kg and the mass of the astronaut is M = 67.0 kg. Before the collision, the bagged lunch moves with a velocity of vi = 1.60 m/s in the positive x-direction and the astronaut is at rest relative to the space station with velocity Vi = 0.         MVi + mvi = MVf + mvf to obtain (67.0 kg)(0 m/s) + (1.40 kg)(1.60 m/s) = (67.0 kg)Vf + (1.40 kg)vf. Because vf = Vf, this further simplifies to (67.0 kg)(0 m/s) + (1.40 kg)(1.60 m/s) = (1.40 kg + 67.0 kg)Vf.            What is the velocity of the astronaut after the collision?