2021 A/L Physics Past Paper · Question 06 (English Medium)

Answer: (4) $\dfrac{3}{2}v$

Given information

We know:

The first block has mass $M$ and velocity $2v$.

The second block has mass $M$ and velocity $v$ in the same direction.

The collision is perfectly inelastic, so the blocks move together afterwards.

What the question asks

We want: the velocity $V$ of the first block after the collision.

The blocks move together, so this is also the velocity of the second block.

Relevant theory

For a system of interacting objects, the momentum remains constant, provided no external force acts on the system.

momentum = mass × velocity

Required equations

$p_{\text{before}} = M(2v) + M(v)$

$p_{\text{after}} = (M + M)V$

$p_{\text{before}} = p_{\text{after}}$

Applying it to this question

Every velocity is in the same unit.

So $V$ comes out in the same unit as $v$.

$M(2v) + M(v) = (M + M)V$

$3Mv = 2MV$

$1.5$

$V = \dfrac{3}{2}v$

Reasoning and calculation

The total momentum before is $3Mv$.

It stays $3Mv$ after, shared by a mass of $2M$.

So the velocity is $\dfrac{3Mv}{2M} = \dfrac{3}{2}v$.

Option (4).

Checks

$(M * a + M * b) / (M + M)$

Does it make sense?

The common velocity must lie between $v$ and $2v$.

$\dfrac{3}{2}v$ is exactly halfway. That is right, because the masses are equal.

Conclusion

$V = \dfrac{M(2v) + M(v)}{M + M} = \dfrac{3}{2}v$

The blocks move together at $V = \dfrac{3}{2}v$.

The answer is (4).

In a perfectly inelastic collision the objects move together after it.

Use conservation of momentum with one common velocity.

If the masses are equal, that velocity is the average.

Exam tip