As shown in the figure, block A of mass m starts from rest from the top of the wedge of the mass 9 m . Wedge can slide on smooth ground. Block A slides down onto the ground, moves along it with a spee

Question

As shown in the figure, block A of mass m starts from rest from the top of the wedge of the mass 9 m . Wedge can slide on smooth ground. Block A slides down onto the ground, moves along it with a speed u, has an elastic collision with the wall, and climbs back onto the wedge. All surfaces are frictionless. Choose the wrong alternative.

TestHub · Accepted Answer

Until the wall is hit, there is no external horizontal impulse, so the center of mass (COM) stays at rest. Thus, (A) is true.

An elastic wall bounce sends A from $+ u$ to $- u$ . The wall's impulse, $2 m u$ , gives the system's COM a speed of:

$\frac{2 m u}{10 m} = \frac{u}{5}$

Therefore, (B) is true.

At maximum height, the block and wedge move together with the COM speed of $\frac{u}{5}$ . Thus, (C) is true.

Let's perform an energy check:

$m g h = \frac{1}{2} m u^{2} + \frac{1}{2} (9 m) (\frac{u}{9})^{2} = \frac{5}{9} m u^{2}$

The final kinetic energy (KE) is:

$KE = \frac{1}{2} (10 m) (\frac{u}{5})^{2} = \frac{m u ^{2}}{5} = \frac{9}{25} m g h$

So, $h^{'} = \frac{16}{25} h \neq = h$ . Therefore, (D) is wrong.

The answer is (D).

Physics - Center of Mass Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Advanced 2028 (Class 11)