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Chapter 4: Problem 473

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

Short Answer

Expert verified
The short answer is: \(v=\frac{M + m}{m}\sqrt{2gh}\).

Step by step solution

01

Applying the conservation of momentum

Before the collision, the momentum of the sphere is \(m \cdot v\) and the momentum of the sandbag is 0. After the collision, the momentum of the sphere is 0 and the momentum of the sandbag is \((M + m) \cdot v_{f}\). According to the conservation of momentum principle: \[m \cdot v = (M + m) \cdot v_{f}\] Now, we need to find the final velocity \(v_{f}\) of the sandbag and sphere after the collision.
02

Applying the conservation of energy

After the collision, the sphere and sandbag move together at final velocity \(v_{f}\) and rise to a height h. The potential energy at height h is \((M+m)gh\). Due to the conservation of energy principle: \[\frac{1}{2}(M+m)v_{f}^2 = (M+m)gh\] Next, we need to isolate \(v_f\) from the equation above.
03

Solving for the final velocity

To find the final velocity \(v_{f}\), divide both sides of the energy conservation equation by \((M + m)\): \[\frac{1}{2}v_{f}^2 = gh\] Now, take the square root of both sides to get \(v_{f}\): \[v_{f} = \sqrt{2gh}\]
04

Solving for the initial velocity of the sphere

Substitute \(v_{f}\) from Step 3 into the conservation of momentum equation from Step 1: \[m \cdot v = (M + m) \cdot \sqrt{2gh}\] Finally, solve for the initial velocity v of the sphere: \[v=\frac{M + m}{m}\sqrt{2gh}\] The correct answer corresponds to the option (A).

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