Vector components - change in momentum

In summary, when a ball of mass m bounces off the rail of a pool table, its y-component of velocity is reversed but its x-component remains unchanged. The change in its linear momentum can be calculated using the equation DP = mvcos(a)(-j^) - mvcos(a))+j^), with negative values indicating a change in direction.
  • #1

Homework Statement

An overhead view of the path taken by a ball of mass m as it bounces from the rail of a pool table. The ball's initial speed is v and the angle is a. The bounce reverses the y component of the ball's velocity but does not alter the x component. ... (b) What is the change in the ball's linear momentum in unit - vector notation?

Homework Equations

Change in momentum (DP)
DP(J) = mvcos(a)(-j^) - mvcos(a))+j^)
DP(I) = mvsin(a)(-i^) - mvsin(a))+i^)

The Attempt at a Solution

I know that the initial and final angles are the same. I thought I needed to use a form like this to get the answer

DP(I) + DP (J)

Both of my answers are negative. I'm not sure if that's the problem or if I have my sin and cos for j and i mixed up.
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  • #2
If angle a is measured in the normal way (anti-clockwise from horizontal) then yes, you have your sines and cosines the wrong way round. You can verify this yourself by drawing a right-angled triangle with the momentum vector as the hypotenuse and the components as the adjacent and opposite sides.

If you know that the x-component remains unchanged, then there is no need to consider it in your calculations since motion in the y-direction is independent of motion in the x-direction. Instead, just consider the change in the y-component of momentum.
  • #3

I would first clarify that the homework problem seems to be describing the physical concept of conservation of momentum. In this scenario, the ball's momentum is changing due to the bounce, but the total momentum of the ball and the pool table remains constant.

To answer the question, the change in momentum can be represented in unit-vector notation as DP = m(vcos(a)i^ - vcos(a)i^) + m(vsin(a)j^ - (-vsin(a))j^). This simplifies to DP = 2mvsin(a)j^, indicating that the change in momentum only occurs in the vertical direction (j^) and is dependent on the mass of the ball, its initial speed, and the angle of the bounce. The negative sign indicates that the ball's momentum is in the opposite direction after the bounce.

What are vector components?

Vector components refer to the individual parts of a vector, which is a quantity that has both magnitude (size) and direction. A vector can be broken down into two or three components, depending on whether it is in two or three dimensions.

What is momentum?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

How do vector components affect change in momentum?

Vector components affect change in momentum by determining the direction and magnitude of the change. If the components of a vector are increased, the change in momentum will also increase. Similarly, if the components are decreased, the change in momentum will decrease.

What is the equation for calculating change in momentum?

The equation for change in momentum is Δp = mΔv, where Δp is the change in momentum, m is the mass of the object, and Δv is the change in velocity.

How do you calculate vector components of momentum?

To calculate the vector components of momentum, you can use trigonometric functions and the Pythagorean theorem. The x-component is equal to pcosθ, where p is the magnitude of the momentum and θ is the angle it makes with the x-axis. The y-component is equal to psinθ. In three dimensions, the z-component can be calculated using pz = pcosφ, where φ is the angle between the momentum and the z-axis.

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