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Question

A 100 gram block is loaded onto a spring along a horizontal frictionless surface. The spring has a constant of 800 N/m and is compressed by a length of 35cm. The block is then released so it collides and sticks to a 300 gram block, sliding together.

a) Find the blocks speed just after it loses contact with the spring, in m/s.

b) Find the blocks speed once they collide and stick together, in m/s.

c) If this collision lasts 0.05 seconds, what is the average force between the blocks, in N?

d) If the block were initially at the same temperature and are made of bismuth (c = 123 J/(kg C)), what is the temperature increase, in C? Assume all lost energy went to increasing the blocks thermal energy.​

Answer 1

First you need to recognize the parts of this scenario. I would draw a diagram showing the spring compressed as condition 1, condition 2 is the spring relaxed and the block moving with a speed of v_launched, then condition 3 is a moment before the collision where the 100 g mass has the same speed v_launched and the 300 g mass has a speed of 0 m/s, condition 4 has both blocks stuck together with a speed of v_final.

1. Part A: This is a traditional energy conservation situation, where elastic potential energy is transformed into kinetic energy of the block. You can find the elastic potential energy of condition 1 with the spring constant and compression amount (make sure to convert from cm to m) and then equate that to the kinetic energy, which you can use to solve for the v_launched at condition 2 requested in part a using the 100 g mass (convert this to kg).
2. Part B: There is a collision and all collisions should start with analyzing conservation of momentum. The only object with momentum before the collision is the 100 g mass, so determine the initial momentum based on the 100 g mass and the v_launched. Set that equal to the momentum of the combined masses times the v_final you will be solving for in part b.
3. Part C: during any collision both objects experience equal and opposite impulse found by FΔt = Δp. You can find the change of momentum of the 100 g block easily enough (you have the mass and both v_launched and v_final), which will be the Δp in the equation and the t is given, so solve for F.
4. Part D: requires you to find the loss in kinetic energy, which always happens in an inelastic collision, and use that as the thermal energy added to the bismuth. Find the amount of energy by calculating the initial kinetic energy (only the 100 g mass is moving - make sure to use the mass in kg) and then kinetic energy after (just add the 2 masses in kg and use the v_final). The difference is the energy lost to thermal energy, which will be the Q in Q=mcΔT and you can solve for ΔT.