Question

physics - lesson 17_conservation of momentum problem - solving
total momentum before = total momentum after
$p_{1\\ initial}+p_{2\\ initial}=p_{1\\ final}+p_{2\\ final}$
(initial) $m_1v_1 + m_2v_2=m_1v_1 + m_2v_2$ (final)
$m_1v_1 + m_2v_2=(m_1 + m_2)v_{final}$

1. two suvs are traveling on the same road. suv a has a mass of 800 kg while suv b has a mass of 650 kg. suv a stops at a stop sign. suv b is traveling behind suv a with a momentum of 16250 kg m/s when it hits the back of suv a. after the collision, suv b has a momentum of 6500 kg m/s and suv a is forced forward. how much momentum is transferred from suv b in order to force suv a forward?
2. a 4.0 - kg boy on a bike has an initial velocity of +10 m/s and collides with a 6.0 - kg girl that is stationary. after the collision, both children are stuck together. what is their final velocity?
3. givens/want:
4. equation:
5. substitution:
6. solve + units:

Explanation:

Step1: Identify the principle

Use conservation of momentum.

Step2: Solve for question 1

The momentum transferred from SUV B to SUV A is equal to the final - momentum of SUV A. So the momentum transferred is 6500 kg m/s.

Step3: Solve for question 2

1. Givens/Want:

• \(m_1 = 4.0\ kg\), \(v_1=10\ m/s\), \(m_2 = 6.0\ kg\), \(v_2 = 0\ m/s\), want \(v_{final}\).

1. Equation:

• According to the conservation of momentum \(m_1v_1+m_2v_2=(m_1 + m_2)v_{final}\).

1. Substitution:

• Substitute the values into the equation: \((4.0\ kg\times10\ m/s)+(6.0\ kg\times0\ m/s)=(4.0\ kg + 6.0\ kg)v_{final}\), which simplifies to \(40\ kg\cdot m/s=(10\ kg)v_{final}\).

1. Solve + Units:

• Solve for \(v_{final}\): \(v_{final}=\frac{40\ kg\cdot m/s}{10\ kg}=4\ m/s\).

Answer:

1. 6500 kg m/s
2. 4 m/s