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Chapter 7: Problem 24

A billiard ball of mass \(m=0.250 \mathrm{~kg}\) hits the cushion of a billiard table at an angle of \(\theta_{1}=60.0^{\circ}\) at a speed of \(v_{1}=27.0 \mathrm{~m} / \mathrm{s}\) It bounces off at an angle of \(\theta_{2}=71.0^{\circ}\) and a speed of \(v_{2}=10.0 \mathrm{~m} / \mathrm{s}\). a) What is the magnitude of the change in momentum of the billiard ball? b) In which direction does the change of momentum vector point?

Short Answer

Expert verified
a) The magnitude of the change in momentum of the billiard ball is 4.315 kg.m/s. b) The direction of the change in momentum vector points at an angle of 54.1 degrees from the negative x-axis.

Step by step solution

01

Calculate the initial momentum components

Using the given information about the incoming angle and the speed of the billiard ball, we will find the components of the initial momentum (\(p_{1x}\) and \(p_{1y}\)) by using these equations: $$p_{1x} = m*v_{1}*cos(\theta_1)$$ $$p_{1y} = m*v_{1}*sin(\theta_1)$$ Plugging in the given values, $p_{1x} = 0.250 * 27.0 * cos(60^{\circ}) \\ = 0.250 * 27.0 * 0.5 \\ = 3.375 kg.m/s$ $p_{1y} = 0.250 * 27.0 * sin(60^{\circ}) \\ = 0.250 * 27.0 * \frac{\sqrt{3}}{2} \\ = 5.845 kg.m/s$
02

Calculate the final momentum components

Using the given information about the outgoing angle and speed of the billiard ball, we will find the components of the final momentum (\(p_{2x}\) and \(p_{2y}\)) by using these equations: $$p_{2x} = m*v_{2}*cos(\theta_2)$$ $$p_{2y} = m*v_{2}*sin(\theta_2)$$ Plugging in the given values, $p_{2x} = 0.250 * 10.0 * cos(71^{\circ}) \\ = 0.250 * 10.0 * 0.335 \\ = 0.837 kg.m/s$ $p_{2y} = 0.250 * 10.0 * sin(71^{\circ}) \\ = 0.250 * 10.0 * 0.942 \\ = 2.355 kg.m/s$
03

Calculate the change in momentum components

To find the change in momentum components, we will subtract the initial components from the final components: $$\Delta p_x = p_{2x} - p_{1x}$$ $$\Delta p_y = p_{2y} - p_{1y}$$ $\Delta p_x = 0.837 - 3.375 \\ = -2.538 kg.m/s$ $\Delta p_y = 2.355 - 5.845 \\ = -3.490 kg.m/s$
04

Calculate the magnitude of the change in momentum

We will use the Pythagorean theorem to find the magnitude of the change in momentum: $$|\Delta p| = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}$$ $|\Delta p| = \sqrt{(-2.538)^2 + (-3.490)^2} \\ = \sqrt{6.433 + 12.180} \\ = \sqrt{18.613} \\ = 4.315 kg.m/s$ So, the magnitude of the change in momentum is 4.315 kg.m/s.
05

Calculate the direction of the change in momentum

We will find the angle of the change in momentum vector (\(\alpha\)) by using the arctangent function: $$\alpha = arctan\left(\frac{\Delta p_y}{\Delta p_x}\right)$$ $\alpha = arctan\left(\frac{-3.490}{-2.538}\right) \\ = arctan(1.375) \\ = 54.1^{\circ}$ So, the change in momentum vector points in a direction of \(54.1^{\circ}\) from the negative x-axis. To summarize: a) The magnitude of the change in momentum of the billiard ball is 4.315 kg.m/s. b) The change of momentum vector points in a direction of \(54.1^{\circ}\) from the negative x-axis.

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Most popular questions from this chapter

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