1.6 Two trolleys, $$ P $$ and $$ Q $$, of mass $$ M $$ and $$ 2 M $$ respectively are at rest on a frictionless horizontal surface. The trolleys have a compressed spring between them. The spring is released and the trolleys move apart. Which one of the following statements is true? (2) a) $$ P $$ and $$ Q $$ have equal kinetic energy b) The speed of $$ P $$ is less than the speed of $$ Q $$ c) The sum of the final kinetic energies of $$ P $$ and $$ Q $$ is zero d) The sum of the final momentum of $$ P $$ and $$ Q $$ is zero

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The problem involves conservation of momentum. Initially, the total momentum is zero because both trolleys $$P$$ (with mass $$M$$) and $$Q$$ (with mass $$2M$$) are at rest. When the spring is released, the trolleys move apart, and by the law of conservation of momentum the sum of the momenta must remain zero.

Let the speeds after release be $$v_P$$ for $$P$$ and $$v_Q$$ for $$Q$$. Conservation of momentum gives:

$$
M v_P - 2M v_Q = 0
$$

Solving for $$v_P$$:

$$
v_P = 2v_Q
$$

This shows that the total momentum remains zero (since one momentum is positive and the other negative when they move in opposite directions), which confirms that option **(d)** is correct.

Thus, the correct statement is:
 
$$
\textbf{(d) The sum of the final momentum of } P \textbf{ and } Q \textbf{ is zero.}
$$
The sum of the final momenta of trolleys $$ P $$ and $$ Q $$ is zero.

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