Blocks and Spring Mechanics Question

  • Thread starter lando45
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  • #1
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When I first read this question I thought it would be simple to solve but apparently it's not, I'm having some difficulty with it.

"Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.40 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.50 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. What are the velocities of both blocks following the collision? (For velocities, use + to mean to the right, - to mean to the left.)"

At first I tried using a simple conservation of momentum equation m1u1 + m2u2 = m1v1 + m2v2 but then I released this doesn't take into account the 1.2J of energy from the spring.

So then I tried solving the question using kinetic energy 0.5mv^2 and I came out with this:

0.5 x 0.4 x 2.4^2 = 1.152J
Total energy = initial KE + 1.2J from spring = 2.352J
0.5m1v1^2 + 0.5m2v2^2 = 2.352J
0.2v1^2 + 0.25v2^2 = 2.352J

So then I am left with two unknowns and no simple way of solving. I then tried combining both the momentum method and the KE method but that doesn't really work because momentum and energy have different units so cannot be used together. Help!
 

Answers and Replies

  • #2
447
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I think you're on the right lines. Maybe you should start by ignoring the spring and work out the combined velocity following the collision. Then imagine you had the two blocks at rest when the spring is released, forcing them apart.
 
  • #3
84
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Ah OK. Well I did that and here's what I got:

m1u1 + m2u2 = (m1 + m2)v
(0.4 x 2.4) + (0.5 x 0) = (0.4 + 0.5)v
0.96 = 0.9v
v = 1.067ms^-1

0.5mv^2 = 1.2J
mv^2 = 2.4J
v^2 = 2.4 / 0.9 = 2.67
v = 1.633ms^-1

Combined velocity = (1.067 + 1.633)ms^-1 = 2.7ms^-1


So then do I just split this combined velocity according to the same ratio of the mass's? i.e. 2.7 = 5x + 4x so the final velocity of m1 is 1.5ms^-1 and the final velocity of m2 is 1.2ms^-1?
 

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