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Chapter 7: Problem 58

Body A of mass \(M\) has an original velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in the \(+x\)-direction toward a stationary body (body \(\mathrm{B}\) ) of the same mass. After the collision, body A has velocity components of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction and \)2.0 \mathrm{m} / \mathrm{s}$ in the + y-direction. What is the magnitude of body B's velocity after the collision?

Short Answer

Expert verified
Answer: The magnitude of the velocity of body B after the collision is approximately 5.39 m/s.

Step by step solution

01

Set up the momentum conservation equations for x and y directions

We know that momentum is conserved in each direction during the collision. Let's set up the momentum equations for the x and y directions separately: In x-direction: Before collision: \(M \cdot v_{Ax}=M \cdot v_{Ax(init)} = M\cdot6.0\) After collision: \(M (v_{Ax(final)}+v_{Bx}) = M(1.0+v_{Bx})\) In y-direction: Before collision: \(M \cdot 0 = 0\) After collision: \(M (v_{Ay(final)}+v_{By}) = M(2.0+v_{By})\)
02

Solve the momentum conservation equations

We can now solve these momentum conservation equations for the final velocity components of body B, \(v_{Bx}\) and \(v_{By}\). Solve for \(v_{Bx}\): \(M \cdot 6.0 = M(1.0+v_{Bx}) \implies 6.0 = 1.0+v_{Bx} \implies v_{Bx} = 5.0 \: m/s\) Solve for \(v_{By}\): \(M \cdot 0 = M(2.0+v_{By}) \implies 0 = 2.0+v_{By} \implies v_{By} = -2.0 \: m/s\)
03

Find the magnitude of the final velocity of body B

Now that we have found the x and y components of the final velocity of body B, we can find the magnitude of its final velocity using the Pythagorean theorem: \(V_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{(5.0)^2 + (-2.0)^2} = \sqrt{25+4} = \sqrt{29} \approx 5.39 \, m/s\) The magnitude of the velocity of body B after the collision is approximately \(5.39 \, m/s\).

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Most popular questions from this chapter

A BMW of mass \(2.0 \times 10^{3} \mathrm{kg}\) is traveling at $42 \mathrm{m} / \mathrm{s} .\( It approaches a \)1.0 \times 10^{3} \mathrm{kg}$ Volkswagen going \(25 \mathrm{m} / \mathrm{s}\) in the same direction and strikes it in the rear. Neither driver applies the brakes. Neglect the relatively small frictional forces on the cars due to the road and due to air resistance. (a) If the collision slows the BMW down to \(33 \mathrm{m} / \mathrm{s},\) what is the speed of the \(\mathrm{VW}\) after the collision? (b) During the collision, which car exerts a larger force on the other, or are the forces equal in magnitude? Explain.
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