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Chapter 7: Problem 63

A projectile of mass \(2.0 \mathrm{kg}\) approaches a stationary target body at \(5.0 \mathrm{m} / \mathrm{s} .\) The projectile is deflected through an angle of \(60.0^{\circ}\) and its speed after the collision is $3.0 \mathrm{m} / \mathrm{s}$ What is the magnitude of the momentum of the target body after the collision?

Short Answer

Expert verified
Answer: 8.7 kg·m/s

Step by step solution

01

Calculate the initial momentum of the projectile

To find the initial momentum of the projectile, we use the formula: momentum = mass × velocity. Given the mass of the projectile is \(2.0\,\text{kg}\) and its initial velocity is \(5.0\,\text{m/s}\), the initial momentum of the projectile is: \(p_{proj init} = m_{proj} \times v_{proj init} = 2.0\,\text{kg} \times 5.0\,\text{m/s} = 10 \,\text{kg}\cdot\text{m/s}\)
02

Split the final velocity of the projectile into X and Y components

After the collision, the projectile is deflected through an angle of \(60^\circ\). To find the magnitude of the momentum of the target body after the collision, we'll first need to break down the final velocity of the projectile into its X and Y components, using trigonometry: \(v_{proj final_x} = v_{proj final} \times \cos(60.0^\circ) = 3.0\,\text{m/s} \times \frac{1}{2} = 1.5 \,\text{m/s}\) \(v_{proj final_y} = v_{proj final} \times \sin(60^\circ) = 3.0\,\text{m/s} \times \frac{\sqrt{3}}{2} = 2.6\,\text{m/s}\)
03

Calculate the final momentum of the projectile

Now we calculate the final momentum of the projectile in the X and Y directions: \(p_{proj final_x} = m_{proj} \times v_{proj final_x} = 2.0\,\text{kg} \times 1.5\,\text{m/s} = 3.0\,\text{kg}\cdot\text{m/s}\) \(p_{proj final_y} = m_{proj} \times v_{proj final_y} = 2.0\,\text{kg} \times 2.6\,\text{m/s} = 5.2\,\text{kg}\cdot\text{m/s}\)
04

Apply conservation of momentum to find the target body's final momentum

Since momentum must be conserved in both the X and Y directions, we can now find the final momentum of the target body: \(p_{target final_x} = p_{proj init} - p_{proj final_x} = 10\,\text{kg}\cdot\text{m/s} - 3.0\,\text{kg}\cdot\text{m/s} = 7.0 \,\text{kg}\cdot\text{m/s}\) \(p_{target final_y} = 0 - p_{proj final_y} = -5.2\,\text{kg}\cdot\text{m/s}\)
05

Calculate the magnitude of the momentum of the target body after the collision

Finally, we can calculate the magnitude of the momentum of the target body after the collision, using the Pythagorean theorem: \(p_{target final} = \sqrt{p_{target final_x}^2 + p_{target final_y}^2} = \sqrt{(7.0\,\text{kg}\cdot\text{m/s})^2 + (-5.2\,\text{kg}\cdot\text{m/s})^2} = 8.7\,\text{kg}\cdot\text{m/s}\) Thus, the magnitude of the momentum of the target body after the collision is \(8.7\,\text{kg}\cdot\text{m/s}\).

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Most popular questions from this chapter

the ball of mass \(145 \mathrm{g},\) and hits it so that the ball leaves the bat with a speed of \(37 \mathrm{m} / \mathrm{s}\). Assume that the ball is moving horizontally just before and just after the collision with the bat. (a) What is the magnitude of the change in momentum of the ball? (b) What is the impulse delivered to the ball by the bat? (c) If the bat and ball are in contact for \(3.0 \mathrm{ms}\), what is the magnitude of the average force exerted on the ball by the bat?
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A projectile of mass \(2.0 \mathrm{kg}\) approaches a stationary target body at \(8.0 \mathrm{m} / \mathrm{s} .\) The projectile is deflected through an angle of \(90.0^{\circ}\) and its speed after the collision is $6.0 \mathrm{m} / \mathrm{s}$ What is the speed of the target body after the collision if the collision is perfectly elastic?
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