/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Two particles one of mass \(m\) ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 47

Two particles one of mass \(m\) and the other of mass \(2 m\) are projected horizontally towards each other from the same level above the ground with velocities \(10 \mathrm{~m} / \mathrm{s}\) and \(5 \mathrm{~m} / \mathrm{s}\) respectively. They collide in air and stick to each other. The distance from \(A\), where the combined mass finally land is (A) \(40 \mathrm{~m}\) (B) \(20 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(45 \mathrm{~m}\)

Short Answer

Expert verified
The exercise does not provide complete information to solve the problem since the height 'h' is not given to determine the time in the air 't'.

Step by step solution

01

Calculate initial velocities and masses of the particles

The first particle has mass 'm' and initial velocity 10m/s, while the second particle has mass '2m' and initial velocity 5m/s
02

Determine the overall initial velocity of the system

To compute the overall initial velocity, we'll need to find how the two velocities add up to create a single linear velocity after collision. Since both particles approach each other with opposite directions, the overall initial velocity of collision will be the difference between their magnitudes: v_tot = v1 - v2 = 10m/s - 5m/s = 5m/s
03

Determine the total time in air

After collision, the particle will have an initial horizontal velocity and act under the influence of the gravitational force, falling to the ground as a projectile. To find the time spent in the air, we can use the following equation considering the particle falls from height h: h = 0.5 * g * t^2 Where 'h' is height, 'g' is acceleration due to gravity (approximately 9.8m/s^2), and 't' is the time. Here, we are not given the height 'h', so we cannot determine the time in air 't'.
04

Determine the horizontal distance from A

To find the horizontal distance from point A, we need the overall initial velocity and the time in air 't'. Since we don't have 't', we can't calculate the horizontal distance from point A. This exercise does not provide complete information to solve the problem.

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Most popular questions from this chapter

A gun fires a shell and recoils horizontally. If the shell travels along the barrel with speed \(v\), the ratio of speeds with which the gun recoils, if the barrel is (i) horizontal (ii) inclined at an angle of \(30^{\circ}\) with horizontal, is (A) 1 (B) \(\frac{2}{\sqrt{3}}\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(\frac{1}{2}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with speed of \(36 \mathrm{~km} / \mathrm{h}\) has a head-on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after collision, both the balls move together, then the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

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.

A particle of mass \(m\) is placed on top of a smooth hemispherical wedge of mass \(4 m\) at rest. The hemispherical wedge is kept on a frictionless horizontal surface. The particle is given a gentle push. The angular velocity of the particle relative to the wedge if wedge has velocity \(v\) when particle has fallen an angle \(\theta\) with respect to the wedge is (A) \(\frac{5 v}{R \cos \theta}\) (B) \(\frac{v}{R \cos \theta}\) (C) \(\frac{4 v}{R \cos \theta}\) (D) \(\frac{5 v}{R \sin \theta}\)

A particle is moving with constant velocity has initial momentum \(P\) is given an impulse of magnitude \(I\), If there is no change is its kinetic energy of the particle then (A) angle between its initial momentum and impulse must be \(<90^{\circ}\). (B) angle between its initial momentum and impulse must be \(>90^{\circ}\). (C) angle between its initial momentum and impulse is \(90^{\circ}\). (D) Not possible.
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