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I am attacking the given problem (as a preface I'm not asking to be spoon fed any answers, just looking for clarity from people much smarter than myself)

A 15.0kg block is attached to a very light horizontal spring of force constant 525N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00m/s horizontally to the left.

Find the maximum distance that the block will compress the spring after the collision.

I have begun by calculating the velocity immediately after collision. Because of the language, I am assuming that it is an elastic collision and therefore kinetic energy is conserved both conservation of momentum and conservation of kinetic energy should work. (I realize, if I knew Kinetic energy, I wouldn't need to know velocity, but I am just checking both ways for my own sanity)

Conservation of KE:

$$ \left(\frac12\right)\left(m_a\right)\left(v_1\right)^2 + \left(\frac12\right)\left(m_b\right)\left(v_1\right)^2 = \left(\frac12\right)\left(m_a\right)\left(v_2\right)^2 + \left(\frac12\right)\left(m_b\right)\left(v_2\right)^2 $$ $$ \Rightarrow \left(\frac12\right)\left(3\right)\left(8\right)^2 = \left(\frac12\right)\left(3\right)\left(2\right)^2 + \left(\frac12\right)\left(15\right)\left(v_2\right)^2 $$ $$ \Rightarrow V_2 = 3.46 $$

Conservation of Momentum:

$$ \left(m_a\right)\left(v_1\right) + \left(m_b\right)\left(v_1\right) = \left(m_a\right)\left(v_2\right) + \left(m_b\right)\left(v_2\right) $$

$$ \Rightarrow \left(3\right)\left(8\right) = \left(3\right)\left(2\right) + \left(15\right)\left(v_2\right) $$ $$ \Rightarrow V_2 = 1.2 $$

I've also tried the formula

$$ V_b2 = \left(2m_a*v_a1\right)/\left(m_a+m_b\right) \Rightarrow V_b2 = 2.66 $$

How is it that two different approaches yields different answers? Is my initial assumption about it being an elastic collision?

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  • $\begingroup$ The collision will be elastic, but once the mass starts moving it will transfer its kinetic energy to the potential energy of the spring... $\endgroup$ – danimal Mar 22 '15 at 2:17
  • $\begingroup$ You'll need to use momentum conservation to get the velocity of the block, and then total energy conservation to get the distance moved via $1/2 kx^2$ $\endgroup$ – danimal Mar 22 '15 at 2:20
  • $\begingroup$ And your momenta need to include negative signs if the velocity is in the opposite direction $\endgroup$ – danimal Mar 22 '15 at 2:22
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I shall assume that the spring is to the right of the block and therefore does not get hit, and will take the positive direction to be to the right. The surface is smooth and the spring is ideal so there are no dissipative forces, so all energy involved will be either in kinetic energy or the potential energy of the spring.

To calculate the velocity of the block, mass $M$ say, we use conservation of momentum in the collision with the stone, mass $m$ - the stone comes in at velocity $u_i$ (which is to the right) and leaves with a velocity $u_f$ (which happens to be to the left, but this will only be important when we put the numbers in and will be negative). We'll call the velocity of the block after the collision $v$ (it was at rest before). So total momentum before = total momentum afterwards gives:

$$mu_i = mu_f + Mv$$ so that $$v = {m(u_i-u_f)\over M} = {3(8-(-2))\over 15} = 2m/s.$$ Notice that I have used a negative value for $u_f$ because we are told that the stone leaves to the left, i.e. the negative direction, and that $v$ ends up being positive, i.e. the block moves to the right, the positive direction.

After this collision we are done with the stone, and concentrate on the block and the spring. When the block first comes to rest it will be because all of its kinetic energy has been used to compress the spring, giving the spring potential energy. The distance it moves will correspond to the extension $x$ (or compression if you like) of the spring when it has this value of the potential energy. So KE of the block = PE of the spring gives:

$${1\over 2}Mv^2 = {1\over 2}kx^2$$

so that

$$x^2 = {Mv^2\over k} = {M{\left(m(u_i-u_f)\over M\right)}^2\over k} = {m^2(u_i-u_f)^2\over Mk} = {3^2(8-(-2))^2\over15\times525} = {900\over7875}$$ and so $$x=\sqrt{{900\over7875}} \simeq 0.338m.$$

In the last block of algebra I could have just put in $v=2$ rather than putting the full expression in terms of the given variables, but it would have come out exactly the same, and we can see a little more of the dependence this way.

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