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Chapter 7: Problem 49

A projectile of 1.0-kg mass approaches a stationary body of \(5.0 \mathrm{kg}\) at \(10.0 \mathrm{m} / \mathrm{s}\) and, after colliding, rebounds in the reverse direction along the same line with a speed of $5.0 \mathrm{m} / \mathrm{s} .$ What is the speed of the 5.0 -kg body after the collision?

Short Answer

Expert verified
Answer: The final velocity of the 5.0-kg body after the collision is 3.0 m/s.

Step by step solution

01

Write the linear momentum conservation equation

The linear momentum conservation equation states that the total momentum before a collision is equal to the total momentum after the collision. In this case, we have two bodies (projectile and stationary body), so we can write the equation as: \(m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\) where \(m_1\) and \(m_2\) are the masses of the projectile and the stationary body respectively, \(v_1\) and \(v_2\) are their initial velocities, and \(v_1'\) and \(v_2'\) are their final velocities after the collision.
02

Input the known values

Now, let's plug in the values given in the exercise: \(m_1=1.0\,\mathrm{kg}\), \(v_1=10.0\,\mathrm{m/s}\), \(m_2=5.0\,\mathrm{kg}\), \(v_2=0\,\mathrm{m/s}\) (stationary), and \(v_1'=-5.0\,\mathrm{m/s}\) (since it rebounds in the reverse direction). We want to find \(v_2'\). \((1.0\,\mathrm{kg})(10.0\,\mathrm{m/s}) + (5.0\,\mathrm{kg})(0\,\mathrm{m/s}) = (1.0\,\mathrm{kg})(-5.0\,\mathrm{m/s}) + (5.0\,\mathrm{kg})v_2'\)
03

Solve for \(v_2'\)

We will now solve for \(v_2'\) in the equation we derived in Step 2: \(10\,\mathrm{kg\cdot m/s} = -5\,\mathrm{kg\cdot m/s} + 5.0\,\mathrm{kg}\cdot v_2'\) Now, add \(5\,\mathrm{kg\cdot m/s}\) to both sides of the equation: \(15\,\mathrm{kg\cdot m/s} = 5.0\,\mathrm{kg}\cdot v_2'\) Finally, divide both sides by \(5.0\,\mathrm{kg}\) to get the final velocity \(v_2'\): \(v_2' = \frac{15\,\mathrm{kg\cdot m/s}}{5.0\,\mathrm{kg}}\) \(v_2' = 3.0\,\mathrm{m/s}\) The final velocity of the 5.0-kg body after the collision is \(3.0\,\mathrm{m/s}\).

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