Math Problem Statement

A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of θ1 = 50.0° at a speed of v1 = 2.60 m/s. It bounces off at an angle of θ2 = 59.0° and a speed of v2 = 2.10 m/s.

a) What is the magnitude of the change in momentum of the billiard ball?

Solution

To find the magnitude of the change in momentum, we need to calculate the momentum before and after the collision, and then find the vector difference between them.

Momentum Before and After Collision

The momentum p\vec{p} is given by: p=mv\vec{p} = m \vec{v} where mm is the mass of the object and v\vec{v} is the velocity vector. Since we have angles, we will decompose the velocities into their xx- and yy-components.

Given:

  • Mass, m=0.150kgm = 0.150 \, \text{kg}
  • Initial speed, v1=2.60m/sv_1 = 2.60 \, \text{m/s}
  • Initial angle, θ1=50.0\theta_1 = 50.0^\circ
  • Final speed, v2=2.10m/sv_2 = 2.10 \, \text{m/s}
  • Final angle, θ2=59.0\theta_2 = 59.0^\circ

1. Calculate Initial Momentum Components

The initial momentum components (p1xp_{1x} and p1yp_{1y}) are: p1x=mv1cos(θ1)p_{1x} = m v_1 \cos(\theta_1) p1y=mv1sin(θ1)p_{1y} = m v_1 \sin(\theta_1) Substitute the known values: p1x=0.150×2.60×cos(50.0)p_{1x} = 0.150 \times 2.60 \times \cos(50.0^\circ) p1y=0.150×2.60×sin(50.0)p_{1y} = 0.150 \times 2.60 \times \sin(50.0^\circ)

2. Calculate Final Momentum Components

Similarly, for the final momentum components (p2xp_{2x} and p2yp_{2y}): p2x=mv2cos(θ2)p_{2x} = m v_2 \cos(\theta_2) p2y=mv2sin(θ2)p_{2y} = m v_2 \sin(\theta_2) Substitute the values: p2x=0.150×2.10×cos(59.0)p_{2x} = 0.150 \times 2.10 \times \cos(59.0^\circ) p2y=0.150×2.10×sin(59.0)p_{2y} = 0.150 \times 2.10 \times \sin(59.0^\circ)

3. Change in Momentum Components

The change in momentum components are: Δpx=p2xp1x\Delta p_x = p_{2x} - p_{1x} Δpy=p2yp1y\Delta p_y = p_{2y} - p_{1y}

4. Magnitude of the Change in Momentum

The magnitude of the change in momentum is: Δp=(Δpx)2+(Δpy)2|\Delta \vec{p}| = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}

I'll calculate these values next.The magnitude of the change in momentum of the billiard ball is approximately 0.093kgm/s0.093 \, \text{kg} \cdot \text{m/s}.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

  1. How would the result change if the angles were different?
  2. What happens to the momentum if the ball's mass increases?
  3. How do you calculate the direction of the change in momentum?
  4. How is momentum conserved in such a collision?
  5. What is the significance of the ball's final speed being lower than the initial speed?

Tip: When dealing with momentum changes in two dimensions, always break the momentum into components to make vector calculations easier.

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Math Problem Analysis

Mathematical Concepts

Physics
Vector Decomposition
Momentum

Formulas

Momentum: p = mv
Vector components: px = mv * cos(θ), py = mv * sin(θ)
Change in momentum: Δp = sqrt((Δpx)^2 + (Δpy)^2)

Theorems

Conservation of Momentum in Two Dimensions

Suitable Grade Level

Grades 11-12