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Chapter 5: Problem 65

A particle of mass \(m\) and velocity \(\vec{v}\) collides elastically with a stationary particle of same mass \(m\). If the collision is oblique, then the angle between the velocity vectors of the two particles after the collision is (A) \(\frac{\pi}{4}\) (B) \(\frac{\pi}{3}\) (C) \(\frac{\pi}{2}\) (D) \(\pi\)

Short Answer

Expert verified
The angle between the velocity vectors of the two particles after the collision is (C) \(\frac{\pi}{2}\).

Step by step solution

01

Conservation of momentum

Since momentum is conserved, the initial momentum of the system should equal the final momentum of the system. Mathematically, this can be represented as: \(m\vec{v} = m\vec{v_1} + m\vec{v_2}\) where \(\vec{v_1}\) and \(\vec{v_2}\) are the final velocity vectors of the first and second particles, respectively.
02

Conservation of kinetic energy

Since the collision is elastic, the kinetic energy is conserved. We can write this as: \(\frac{1}{2} m v^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2\) Dividing both sides by \(m/2\), we get: \(v^2 = v_1^2 + v_2^2\)
03

Solve for the angle

Using the equation from Step 1, we can substitute the magnitudes of the velocity vectors to find the cosine of the angle between them: \(v = v_1 + v_2 \cos\theta\) Now, using the equation from Step 2, we square the equation above, which leads to: \(v^2 = v_1^2 + 2v_1v_2\cos\theta + v_2^2\) Notice that this is similar to our equation from Step 2. Subtracting the equation from Step 2 leaves us with: \(2v_1v_2\cos\theta = 0\) Since \(v_1\) and \(v_2\) are non-zero (because it's an oblique collision), it means that: \(\cos\theta = 0\) Therefore, the angle between the velocity vectors of the two particles after the collision is: \(\theta = \frac{\pi}{2}\). So, the correct answer is (C) \(\frac{\pi}{2}\).

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Most popular questions from this chapter

A ball of mass \(m\) approaches a wall of mass \(M(\gg>m)\) with speed \(4 \mathrm{~m} / \mathrm{s}\) along the normal to the wall. The speed of wall is \(1 \mathrm{~m} / \mathrm{s}\) towards the ball. The speed of the ball after an elastic collision with the wall is (A) \(5 \mathrm{~m} / \mathrm{s}\) away from the wall. (B) \(9 \mathrm{~m} / \mathrm{s}\) away from the wall. (C) \(3 \mathrm{~m} / \mathrm{s}\) away from the wall. (D) \(6 \mathrm{~m} / \mathrm{s}\) away from the wall.

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 force of \(-F \hat{k}\) acts on \(O\), the origin of the coordinate system. The torque about the point \((1,-1)\) is (A) \(-F(\hat{i}-\hat{j})\) (B) \(F(\hat{i}-\hat{j})\) (C) \(-F(\hat{i}+\hat{j})\) (D) \(F(\hat{i}+\hat{j})\)

Two bodies of mass \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) move towards each other in mutually perpendicular direction with the velocities \(3 \mathrm{~m} / \mathrm{s}\) and \(2 \mathrm{~m} / \mathrm{s}\) respectively. If the bodies stick together after collision the energy loss will be (A) \(13 \mathrm{~J}\) (B) \(\frac{13}{3} \mathrm{~J}\) (C) \(8 \mathrm{~J}\) (D) \(7 \mathrm{~J}\)

Two particles of masses \(4 \mathrm{~kg}\) and \(8 \mathrm{~kg}\) are separated by a distance of \(6 \mathrm{~m}\). If they are moving towards each other under the influence of a mutual force of attraction, then the two particles will meet each other at a distance of (A) \(6 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (B) \(2 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (C) \(4 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass (D) \(8 \mathrm{~m}\) from \(8 \mathrm{~kg}\) mass
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