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A small sphere of mass \(m=1 \mathrm{~kg}\) is moving with a veloctin \((4 \hat{i}-\hat{j}) \mathrm{m} / \mathrm{s}\). It hits a fixed smooth wall and rebounds wit velocity \((\hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}\). The coefficient of restitution between the sphere and the wall is \(n / 16 .\) Find the value of \(n\).

Short Answer

Expert verified
The value of \(n\) is 64.

Step by step solution

01

Understand the Coefficient of Restitution Formula

The coefficient of restitution, usually denoted by \(e\), is defined by the formula \( e = \frac{v_2 - v_1}{u_1 - u_2} \), where \( u_1 \) and \( v_1 \) are the initial and final velocities of the object in the direction perpendicular to the surface, and \( u_2 \) and \( v_2 \) are zero since the wall is fixed.
02

Extract Relevant Velocity Components

Since the wall is smooth, only the perpendicular (\( \hat{j} \)) component of the velocity is affected. Thus, the initial perpendicular velocity of the sphere is \( u_1 = -1 \) m/s, and the final perpendicular velocity \( v_1 = 3 \) m/s.
03

Calculate the Coefficient of Restitution

Using the formula \( e = \frac{v_1 - u_1}{u_1} \), we substitute the values: \( e = \frac{3 - (-1)}{-1} = -4 \). However, because the velocities are in opposite directions, \( e \) should be positive: \( e = 4 \).
04

Solve for n

Given that the coefficient of restitution is \( \frac{n}{16} \), we equate and solve for \( n \): \( \frac{n}{16} = 4 \). Therefore, \( n = 64 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collision
An elastic collision is a type of collision where no kinetic energy is lost during the interaction between two objects. This concept is essential when considering collisions in physics because it helps predict how objects will behave post-collision. Kinetic energy and momentum are conserved in elastic collisions. This means that the total energy and momentum before the collision is the same as after.In our exercise, when the sphere hits the smooth wall, it can be treated as a near-elastic collision regarding the perpendicular movement. The sphere has two velocity components, and the component perpendicular to the wall (the \( \hat{j} \) direction) plays a crucial role in determining how the sphere rebounds. Kinetic energy is conserved for motion along the wall. Thus, although some of the sphere's kinetic energy along the \( \hat{j} \) axis is conserved, the nature of the fixed wall dictates the response in the normal direction.
Velocity Components
Velocity is a vector quantity, which means it has both magnitude and direction. When analyzing motions like that of the sphere, we break down the velocity into components, usually along horizontal (\( \hat{i} \)) and vertical (\( \hat{j} \)) directions. This simplifies solving problems of motion, especially when dealing with collisions.In this problem, the sphere's initial velocity before hitting the wall is \((4 \hat{i} - \hat{j})\), meaning it moves both to the right and downward. After rebounding, it's expressed as \((\hat{i} + 3\hat{j})\), showing movement to the right and upward. Notably, the \( \hat{j} \) components indicate how the velocity changes due to the collision, from \(-1\) m/s to \(3\) m/s.The \( \hat{i} \) component, parallel to the wall, remains unchanged as the wall exerts no force in this direction. By breaking down the motion like this, it becomes easier to apply laws of motion and collision, such as momentum conservation and the coefficient of restitution.
Momentum Conservation
Momentum conservation is a fundamental principle stating that the total momentum of a closed system remains constant if no external forces act on it. In the context of the sphere hitting the wall, momentum conservation helps us understand how the velocities change.While momentum is generally conserved in the entire system of objects considered, due to the wall being fixed (it doesn't move), there is no change in its momentum. Instead, we analyze the sphere's momentum change. Before the collision, the sphere's momentum in the \( \hat{j} \) direction was \(-1\) kg⋅m/s (calculated as mass times velocity, \(m \times \hat{j}\)). Post-collision, it becomes \(3\) kg⋅m/s.Because the wall does not exert any horizontal forces in the \( \hat{i} \) direction, the sphere's momentum in this direction remains \(4\) kg⋅m/s. This kind of analysis, using both velocity components and momentum, helps us to derive key quantities like the coefficient of restitution, which measures how 'bouncy' a collision is—essentially quantifying the deviation from a perfectly elastic collision.

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Most popular questions from this chapter

A ball is dropped from a height of \(45 \mathrm{~m}\) from the ground. The coefficient of restitution between the ball and the ground is \(2 / 3\). What is the distance travelled by the ball in 4th second of its motion? Assume negligible time is spent in rebounding. Le \(g=10 \mathrm{~m} \mathrm{~s}^{2}\) (1) \(5 \mathrm{~m}\) (2) \(20 \mathrm{~m}\) (3) \(15 \mathrm{~m}\) (4) \(10 \mathrm{~m}\)

A ball of mass \(m=1 \mathrm{~kg}\) is hung vertically by a thread of length \(l=1.50 \mathrm{~m}\). Upper end of the thread is attached to the ceiling of a trolley of mass \(M=4 \mathrm{~kg}\). Initially, the trolley is stationary and it is free to move along horizontal rails without friction. A shell of mass \(m=1 \mathrm{~kg}\), moving horizontally with velocity \(v_{0}=6 \mathrm{~m} / \mathrm{s}\) collides with the ball and gets stuck with it. As a result, the thread starts to deflect towards right. The velocity of the combined body just after collision is (1) \(2 \mathrm{~m} / \mathrm{s}\) (2) \(3 \mathrm{~m} / \mathrm{s}\) (3) \(\mid \mathrm{m} / \mathrm{s}\) (4) \(4 \mathrm{~m} / \mathrm{s}\)

Following quantities of the ball will remain a constant before and after any collision (1) linear momentum (2) kinetic energy, angular momentum about the centre of the circle (3) velocity, angular momentum about the centre of the circle, kinetic energy (4) none of these

A hockey player receives a corner shot at a speed of 15 \(\mathrm{m} / \mathrm{s}\) at angle \(30^{\circ}\) with \(y\)-axis and then shoots the ball along \(x\)-axis with the speed \(30 \mathrm{~m} / \mathrm{s}\). If the mass of the ball is \(1 \mathrm{sl}\) \(\mathrm{g}\) and it remains in contact with the hockey stick for \(0.01 \mathrm{~s}\) the force exerted on the ball along \(x\)-axis is (1) \(281 \mathrm{~N}\) (2) \(187.5 \mathrm{~N}\) (3) \(562.5 \mathrm{~N}\) (4) \(375 \mathrm{~N}\)

In the figure, the block \(B\) of mass \(m\) starts from rest at the top of a wedge \(W\) of mass \(M\). All surfaces are without friction. \(W^{\prime}\) can slide on the ground. \(B\) slides down onto the ground, moves along it with a speed \(v\), has an elastic collision with the wall, and climbs back on to \(W\). (1) From the beginning, till the collision with the wall, the centre of mass of ' \(B\) plus \(W\) does not move horizontally. (2) After the collision, the centre of mass of ' \(B\) plus \(W\) moves with the velocity \(\frac{2 m v}{m+M}\). (3) When \(B\) reaches its highest position of \(W\), the speed of \(W\) is \(\frac{2 m v}{m+M}\) (4) When \(B\) reaches its highest position of \(W\), the speed of \(W\) is \(\frac{m v}{m+M}\)

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