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Chapter 7: Problem 45

A 2.0-kg block is moving to the right at \(1.0 \mathrm{m} / \mathrm{s}\) just before it strikes and sticks to a 1.0-kg block initially at rest. What is the total momentum of the two blocks after the collision?

Short Answer

Expert verified
Answer: The total momentum of the two blocks after the collision is 2.0 kg m/s.

Step by step solution

01

Express the given values and the conservation of momentum principle

First, express the given values and the momentum formula. Initially, let the mass of block A be \(m_A = 2.0 \,\text{kg}\) and its velocity be \(v_A = 1.0 \,\text{m/s}\). Let the mass of block B be \(m_B = 1.0 \,\text{kg}\) and its initial velocity be \(v_B = 0 \,\text{m/s}\) as it is at rest. The total momentum before collision is the sum of the momenta of individual blocks, i.e., \(p_{\text{total}} = m_A v_A + m_B v_B\).
02

Calculate the total momentum before collision

Use the formula and the given values to calculate the total momentum before the collision: \(p_{\text{total}} = m_A v_A + m_B v_B = (2.0 \,\text{kg})(1.0 \,\text{m/s}) + (1.0 \,\text{kg})(0 \,\text{m/s}) = 2.0 \,\text{kg m/s}\). The total momentum before the collision is \(2.0 \,\text{kg m/s}\).
03

Apply the conservation of momentum principle

The conservation of momentum principle states that the total momentum before the collision is equal to the total momentum after the collision. Since both blocks stick together after the collision, they move together as one single mass with the same velocity. The combined mass can be written as \(m_{\text{combined}} = m_A + m_B\). Let the velocity of the combined blocks after the collision be \(v_f\). Then the total momentum after the collision can be stated as \(p_{\text{total}} = m_{\text{combined}} v_f\).
04

Calculate the velocity of the combined blocks after the collision

To find the velocity of the combined blocks after the collision, set the total momentum before the collision equal to the total momentum after the collision and solve for the velocity \(v_f\): \(2.0 \,\text{kg m/s} = (2.0 \,\text{kg} + 1.0 \,\text{kg}) v_f\). Now, divide both sides by the combined mass: \(v_f = \frac{2.0 \,\text{kg m/s}}{3.0 \,\text{kg}} = 0.67\,\text{m/s}\). The velocity of the combined blocks after the collision is \(0.67\,\text{m/s}\).
05

Calculate the total momentum after the collision

Finally, use the velocity of the combined blocks after the collision and the combined mass to find the total momentum after the collision: \(p_{\text{total}} = m_{\text{combined}} v_f = (3.0 \,\text{kg})(0.67\,\text{m/s}) = 2.0\,\text{kg m/s}\). The total momentum of the two blocks after the collision is \(2.0\,\text{kg m/s}\).

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