Question

A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart (Figure 9-E6). The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L. Find the displacement of the cart during this process.

Answer 1

Thanks for asking the question!

ANSWER::

See figure for better explanation!

Mass of bob = m
Mass of cart = M
Let us assume that bob falls at A

Centre of mass in the inital time = (m x L + M x 0) / M+m = mL/(M+m)

Distance from P::

When the bob fall in the gap the centre of mass is at a distance 'O' from P.

Centre of mass shift = 0 - mL/(M+m) = -mL / (M+m) [towards left]

And mL / (M+m) towards right .

And also , there is no external force in horizontal direction .

After this we can conclude that the cart displaces a distance mL / (M+m) towards right.

Hope it helps!

Answer 2

The displacement of the cart during this process is given by $\frac{(mL)}{(M+m)}$.

Explanation:

Step 1:

Bob's mass = m

Cart’s mass = M

Let's suppose bob falls at A

Mass center in the inital time =$\frac{(m \times L+M \times 0)}{M+m}$

                                                $=\frac{(\mathrm{m} \mathrm{L})}{\mathrm{M}+\mathrm{m}}$

Step 2:

Distance from P

The center of mass is at a point ' O ' from P when the bob drops in the gap.

Centre of mass shift $=\frac{0-m L}{(M+m)}$

$=\frac{-m L}{(M+m)}$                 [towards left]

= $\frac{\mathrm{mL}}{(\mathrm{M}+\mathrm{m})}$                   [towards right]

Even, in horizontal direction, there is no external force.  

Then we can assume that the cart is traveling a distance $\frac{(mL)}{((M+m))}$  to the right.