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Chapter 5: Problem 35

Two bodies of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are moving with velocities \(10 \mathrm{~m} / \mathrm{s}\) and \(2 \mathrm{~m} / \mathrm{s}\) towards each other. The velocity of their centre of mass is (A) Zero (B) \(1 \mathrm{~m} / \mathrm{s}\) (C) \(2 \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity of the center of mass for the given system is (C) \(2 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Write down the given information

We have the following given information: - Mass of the first body (\(m_1\)): 2 kg - Mass of the second body (\(m_2\)): 4 kg - Velocity of the first body (\(v_1\)): 10 m/s - Velocity of the second body (\(v_2\)): -2 m/s (opposite direction to \(v_1\))
02

Write the formula for the velocity of the center of mass

The formula for the velocity of the center of mass is: \(V_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}\)
03

Plug in the given values into the formula

Now we insert the given values into the formula: \(V_{cm} = \frac{(2 \mathrm{~kg})(10 \mathrm{~m} / \mathrm{s}) + (4 \mathrm{~kg})(-2 \mathrm{~m} / \mathrm{s})}{2 \mathrm{~kg} + 4 \mathrm{~kg}}\)
04

Perform the calculations

Calculate the expression: \(V_{cm} = \frac{20 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} - 8 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{6 \mathrm{~kg}}\) \(V_{cm} = \frac{12 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{6 \mathrm{~kg}}\) \(V_{cm} = 2 \mathrm{~m} / \mathrm{s}\)
05

Select the correct answer

The calculated velocity of the center of mass is \(2 \mathrm{~m} / \mathrm{s}\). Based on this result, we can confidently choose the correct option. The answer is (C) \(2 \mathrm{~m} / \mathrm{s}\).

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Most popular questions from this chapter

Two blocks of masses \(m_{1}\) and \(m_{2}\) are connected with an ideal spring and kept on a frictionless plane at rest. Another block of mass \(m_{3}\) making elastic head on collision with the block of mass \(m_{1}\). After the collision the centre of mass of \(\left(m_{1}+m_{2}\right)\) as a system will (A) move with non-uniform acceleration. (B) move with a uniform velocity. (C) remain at rest. (D) move with uniform acceleration.

A trolley containing water has total mass \(4000 \mathrm{~kg}\) and is moving at a speed of \(40 \mathrm{~m} / \mathrm{s}\). Now water start coming out of a hole at the bottom of the trolley at the rate of \(8 \mathrm{~kg} / \mathrm{s}\). Speed of trolley after \(50 \mathrm{~s}\) is (A) \(44.44 \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(44 \mathrm{~m} / \mathrm{s}\) (D) \(54.44 \mathrm{~m} / \mathrm{s}\)

A ball after falling through a distance \(h\) collides with an inclined plane of inclination \(\theta\) as shown. It moves horizontally after the impact. The co- efficient of restitution between inclined plane and ball is (A) 1 (B) \(\tan ^{2} \theta\) (C) \(\cot ^{2} \theta\) (D) \(\sin ^{2} \theta\)

A ball weighing \(10 \mathrm{~g}\) hits a hard surface vertically with a speed of \(5 \mathrm{~m} / \mathrm{s}\) and rebounds with the same speed. The ball remains in contact with the surface for \(0.01 \mathrm{~s}\). The average force exerted by the surface on the ball is (A) \(100 \mathrm{~N}\) (B) \(10 \mathrm{~N}\) (C) \(1 \mathrm{~N}\) (D) \(0.1 \mathrm{~N}\)

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