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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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