How Does Momentum Affect Motion on Ice and in Spring-Loaded Systems?

[courtrigrad]

(1) You are standing on a sheet of ice that covers a parking lot; there is negligible friction between your feet and the ice. A friend throws you a 0.400 kg ball that is traveling horizontally at 12.0 m/s. Your mass is 80.0 kg. (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so that afterward it is moving horizontally at 11.0 m/s in the opposite direction, what is your speed after the collision?

(a) So p = mv. That means (0.4 kg)(12.0 \frac{m}{s}) = (80.4 kg)(v) and you just solve for v?
(b) Would you do: (0.4 kg)(11.0 \frac{m}{s}) = (80.0 kg)(v) and solve for v?

(2) Two blocks are forced together, compressing a spring in between them. The first block has a mass of 1.00 kg, and the second block has a mass of 3.00 kg. The spring is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 0.5 m/s. (a) What is the final speed of Block A? (b) How much potential energy was stored in the compressed spring?

(a) So (0.500 m/s)(3.00 kg) = (1.00 kg)(v) and you just solve for v?
(b) The potential energy of a spring is \frac{1}{2}kx^{2}. How would you find k and x?

Thanks

 

[Astronuc]

In problem 1a, if the ball is caught the process is 'inelastic collision' - http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

In the second case, if the ball bounces off, the process is 'elastic collsion' - http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html#c4

An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed.

In 2a, the net momentum is zero since the spring dropped, and the forces must be equal and opposite since the spring did not move.

In 2b, the spring potential energy is converted into kinetic energy, so the total potential energy has to equal the total kinetic energy of both masses.

 

[courtrigrad]

thanks a lot Astronuc