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Chapter 5: Problem 111

Distance travelled by cart before second collision is (A) \(2 L\) (B) \(2 L\left(\frac{m}{M+m}\right)\) (C) \(2 L\left(\frac{M}{M+m}\right)\) (D) \(L+L \frac{m}{M+m}\)

Short Answer

Expert verified
The distance traveled by the cart before the second collision is \(2L \frac{m}{M+m}\), which corresponds to option (B).

Step by step solution

01

Determine the time taken for the mass m to fall freely and collide with the cart for the second time.

From the given problem, we know that the mass M is moving horizontally while the mass m is falling freely. Since the mass m is falling freely, we can use the equation of motion to find the time it takes for the mass m to fall and touch the cart for the second time: \(h = \frac{1}{2}gt^2\) where h is the height the mass m falls, g is the acceleration due to gravity, and t is the time taken for the mass m to fall. In this case, the mass m has to fall a height of 2L to touch the cart for the second time, so we have: \(2L = \frac{1}{2}gt^2\) Solve for t: \(t = \sqrt{\frac{4L}{g}}\)
02

Calculate the acceleration of the cart.

Now we need to find the acceleration of the cart (let's call it a) so that we can calculate the distance it travels before the second collision. The force acting on the system is the weight of the mass m: m * g. Since the pulley is smooth and there is no external force acting on the system, we can use Newton's second law of motion: F = (m + M) * a Solve for a: \(a = \frac{mg}{M+m}\)
03

Calculate the distance traveled by the cart.

We now have the time taken for mass m to touch the cart for the second time and the acceleration of the cart. We can use the equation of motion to find the distance traveled by the cart: Distance = initial_velocity * time + (1/2) * acceleration * time^2 Since the initial_velocity is 0, we can simplify the equation to: Distance = (1/2) * a * t^2 Now, substitute the values of a and t: Distance = (1/2) * \(\frac{mg}{M+m}\) * \(\frac{4L}{g}\) Simplifying the equation, we get: Distance = \(2L \frac{m}{M+m}\) Hence, the distance traveled by the cart before the second collision is \(2L \frac{m}{M+m}\), which corresponds to option (B).

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