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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Centre of Mass, Linear momentum, Collision

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

A cart of mass M rests on a frictionless horizontal surface, and a...

A cart of mass MM rests on a frictionless horizontal surface, and a pendulum bob of mass mm hangs from the roof of the cart as shown in the figure. After the string breaks, the bob falls to the floor, experiences several collisions on the floor, and eventually lands in a small slot made in the floor. The horizontal distance between the string and the slot is LL. Determine the displacement of the cart during this process.
A cart of mass M rests on a frictionless horizontal surface, and a pendulum bob of mass m hangs from the roof of the

Options

  1. AmLM\dfrac{mL}{M}
  2. BMLm+M\dfrac{ML}{m + M}
  3. CmLm+M\dfrac{mL}{m + M}
  4. DMLm\dfrac{ML}{m}

Correct answer

C. mLm+M\dfrac{mL}{m + M}

Step-by-step solution

Since there is no external force acting on the system (cart and bob) in the horizontal direction, the horizontal position of the center of mass of the system remains stationary.

Let xx be the displacement of the cart relative to the ground. The displacement of the bob relative to the cart is LL.

Therefore, the displacement of the bob relative to the ground is x+Lx + L.

Using the principle of conservation of the center of mass position:

MΔxM+mΔxm=0M \Delta x_{M} + m \Delta x_{m} = 0

Mx+m(x+L)=0M x + m (x + L) = 0

(M+m)x+mL=0(M + m) x + mL = 0

x=−mLM+mx = -\dfrac{mL}{M + m}

The negative sign indicates that the cart moves in the direction opposite to the displacement of the bob relative to the cart. The magnitude of the displacement of the cart is mLm+M\dfrac{mL}{m + M}.
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