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Chapter 6: Problem 81

A ball moving with velocity \(v\) and mass \(m\) strikes against a wall at an angle \(\theta\) and is reflected at the same angle, then which of the following is incorrect? (a) \(\Delta|\vec{p}| \neq 0\) (b) \(|\Delta \vec{p}|-0\) (c) \(\Delta K \neq 0\) (d) \(\Delta|\vec{p}|-0\)

Short Answer

Expert verified
The incorrect option is (a) \(\Delta |\vec{p}| \neq 0\).

Step by step solution

01

Understand the Problem

We need to determine which statement is incorrect regarding the change in momentum and kinetic energy (if any) when a ball strikes a wall and reflects at the same angle. - Given: Velocity \(v\), Mass \(m\), Angle \(\theta\).- Outcome: Determine the incorrect option regarding changes in \(\Delta |\vec{p}|\) and \(\Delta K\).
02

Momentum Before and After Impact

Before the impact, the ball's momentum can be expressed in vector form based on its direction. After striking the wall, since it reflects at the same angle, the magnitude remains unchanged, but the direction changes. Thus:- Initial momentum: \(\vec{p}_{i} = m \cdot v\)- Final momentum: \(\vec{p}_{f} = m \cdot v\) (direction reflects).However, since magnitude does not change, \(\Delta |\vec{p}| = 0\). Thus option (a) is incorrect.
03

Analyze Kinetic Energy

Since the kinetic energy depends only on the speed (not the direction) of the ball and there is no speed change:- Initial kinetic energy: \(K_i = \frac{1}{2} m v^2\)- Final kinetic energy: \(K_f = \frac{1}{2} m v^2\).The change in kinetic energy (\(\Delta K\)) is zero as there is no change in speed, affirming option (c) as correct.
04

Conclusion

Evaluating the given options:- (a) Incorrect since \(\Delta |\vec{p}| = 0\), the magnitude does not change.- (b) Seems incorrect, but actually states that the change in the magnitude of momentum is zero which is true, thus correct.- (c) Correct since \(\Delta K = 0\).- (d) Correct since \(\Delta |\vec{p}| = 0\).The incorrect statement is (a) \(\Delta |\vec{p}| eq 0\).

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Key Concepts

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

Change in Momentum
When we talk about the change in momentum, we're dealing with the quantity of motion that an object carries. Momentum is a vector, meaning it has both magnitude and direction. It's calculated as the product of an object's mass and velocity: \[ \vec{p} = m \cdot \vec{v} \]However, reflecting on the wall doesn't alter the magnitude of the ball's speed but changes its direction. This can create a scenario where the momentum's magnitude remains constant, but its direction shifts. - **Before Impact:** The ball approaches the wall with an initial momentum given by the formula above, based on its mass and approach velocity.- **After Impact:** The ball's direction is reversed, but since it's said to reflect at the same angle, the speed鈥攁nd therefore the momentum's magnitude鈥攔emains constant.Thus, the change in the magnitude of the momentum, \(\Delta |\vec{p}|\), is zero, meaning there is no net change in the speed component of momentum. But, remember, the change in direction itself implies a vectorial change, just not a scalar magnitude change.
Kinetic Energy
Kinetic Energy (KE) is the energy that an object possesses due to its motion, depending solely on the speed of the object, not its direction. The formula for kinetic energy is:\[ K = \frac{1}{2} m v^2 \]- **Before Collision:** The ball has an initial kinetic energy which is determined by its mass and velocity squared.- **After Collision:** Since the speed remains unchanged post-collision, the kinetic energy remains the same.Important to note is that even when the ball bounces off the wall and changes direction, as long as the speed is constant, the kinetic energy remains unchanged. Thus, for an elastic collision where there is no loss of speed, \( \Delta K = 0 \).
Elastic Collision
An elastic collision is one where there is no net loss of total kinetic energy in the system. In the scenario provided, the ball striking and reflecting at an angle demonstrates a perfect elastic collision. During elastic collisions: - **Energy Conservation:** Both the total kinetic energy and momentum are conserved. - **No Deformation:** Neither the ball nor the wall have permanent deformations post-collision. - **No Heat Loss:** There's negligible conversion of kinetic energy into other forms like heat. This situation ensures that both speed and kinetic energy remain constant pre and post-collision, maintaining the integrity of an elastic collision scenario.

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Most popular questions from this chapter

Two same masses are tied with equal lengths of strings and are suspended at the same fixed point. One mass is suspended freely whereas another is kept in a way that string is horizontal as shown. This mass is given initial velocity \(u\) in vertical downward direction, it strikes the freely suspended mass elastically that is just able to complete the circular motion after the collision about point of suspension \(O .\) Magnitude of velocity \(u\) is: (a) \(\sqrt{g L}\) (b) \(\sqrt{2.5 g L}\) (c) \(\sqrt{2 g L}\) (d) \(\sqrt{3 g L}\)

A smooth sphere of mass \(m\) is moving on a horizontal plane with a velocity \((3 \hat{i}+\hat{j}) .\) lt collides with a vertical wall which is parallel to the vector \(\hat{j}\). If \(e^{-} \frac{1}{2}\) then impulse \((\vec{J})\) that acts on the sphere is: (a) \(-\frac{9}{2} m \hat{i}\) (b) \(\left(-\frac{3}{2} \hat{i}+\hat{j}\right)\) (c) \(\frac{3}{2} m \hat{j}\) (d) \(\left(\frac{3}{2} m \hat{j}+\frac{1}{2} m \hat{i}\right)\)

\(\Lambda\) block of mass \(m_{1}\) is connected to another block of mass \(m_{2}\) by a light spring of spring constant \(k\). 'The blocks are kept on a smooth horizontal surface. Initially the blocks are at rest and the spring is unstretched when a constant force \(F\) starts acting on the block of mass \(m_{2}\) to pull il. Find the maximum clongation of the spring.

A body \(A\) of mass \(2 \mathrm{~kg}\) is projected upward from the surface of the ground at \(t=0\) with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). One sccond after a body \(B\), also of mass \(2 \mathrm{~kg}\) is dropped from a height o \(20 \mathrm{~m}\). Ir they collide elastically, then their velocities just after collision are: (a) \(v_{A}=5 \mathrm{~m} / \mathrm{s}\) downward, \(v_{\mathrm{B}}=5 \mathrm{~m} / \mathrm{s}\) upward (b) \(v_{\mathrm{A}}=10 \mathrm{~m} / \mathrm{s}\) downward, \(v_{\mathrm{B}}=5 \mathrm{~m} / \mathrm{s}\) upward (c) \(v_{\Lambda}=10 \mathrm{~m} / \mathrm{s}\) upward \(v_{\mathrm{B}}=10 \mathrm{~m} / \mathrm{s}\) downward (d) both move downward with velocity \(5 \mathrm{~m} / \mathrm{s}\)

For the carth-moon system let \(M\) and \(m\) be the masses o the earth and the moon respectively. Let \(\vec{v}\) be the instantaneous relative velocity. 'l'he total kinetic energy of this system in the centre of mass frame will be given by: (a) \(\frac{1}{2} \frac{m M}{(m+M)} v^{2}\) (b) \(\frac{1}{2} m^{2}\) ? \(\frac{1}{2} M v^{2}\) (c) \(\frac{1}{2} m v^{2}\) (d) \(\frac{1}{2} M v^{2}\)
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