A bullet of mass 2.0g is fired horizontally into a block of wood of mass 600g. The block is suspended from strings so that it is free to move in a vertical plane. The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6cm, as shown below:

enter image description here

(a) (i) Calculate the change in gravitational potential energy of the block and bullet.

(b)Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3ms$^{–1}$.

Source:A levels 9702/02/M/J/05

My Approach: I calculate the Potential energy using E=mgh and get E=0.51J. My query is about the second question. They move off together as soon as bullet buries the block.How can I relate 8.6 cm and question required velocity.

The marking scheme says:

$$v^2=u^2+2gh=2\times9.8\times0.086 \qquad[\text{Since }u=0]$$

But how can the v there be the initial speed of the block and the bullet?I kinda don't understand for the equation where the start of motion is taken and till when?

  • $\begingroup$ Are you taking components of speed? If yes then which one and at which instant? $\endgroup$
    – Amar30657
    Sep 13, 2017 at 12:40
  • 1
    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ Sep 13, 2017 at 12:40
  • $\begingroup$ Sorry i actually wanted to understand rather than homework, if i was to do for homework i would have copied from marking scheme. $\endgroup$
    – Amar30657
    Sep 13, 2017 at 12:41
  • $\begingroup$ I think you confused by semantics and are trying to overthink this. The bullet hits the block and, immediately after the collision, the bullet-block system will have some initial velocity, enough to raise the bullet-block to some height $h$. When it reaches $h$, the system will be instantenously at rest. Short of someone solving the problem for you that's should be enough to clarify your issues. $\endgroup$ Sep 13, 2017 at 12:45
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    $\begingroup$ I disagree with the close votes on this question. As seen in the last paragraph, this is clearly asked at the conceptual level, not the "do my homework for me" level. $\endgroup$
    – Floris
    Sep 13, 2017 at 13:53

1 Answer 1


There are three time points to think about:

  1. Just before the bullet A hits the block B: the bullet has some velocity $v_A$ and the block is stationary. The bullet has a certain kinetic energy $\frac12 m_A v_A^2$
  2. Right after impact: the bullet has transferred some momentum to the block, and some energy was dissipated during the impact. They are now moving together at some initial speed $v_i$ (which the marking scheme calls $v$), with height $h=0$. The total kinetic energy is $\frac12 (m_A + m_B) v_i^2$.
  3. At the end of the swing, the bullet plus block have reached some height $h$ and velocity $u=0$; by using the bifilar suspension, we make sure the COM is moving up by this amount regardless of where it is located (angle of block doesn't change). All the kinetic energy from (2) is now turned into potential energy

At any moment between (2) and (3), the sum of kinetic and potential energy is constant.

Does that clear it up for you? Are you just confused about the choice of $u$ and $v$?.

  • $\begingroup$ Maybe the OP is concerned about the time between 1 and 2 - ie when the block has started moving but the bullet has not yet stopped in the block. $\endgroup$ Sep 13, 2017 at 17:30

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