Question

1. a 6-kg bowling ball rolling at 5 m/s strikes a stationary 4-kg bowling ball. if the 6-kg ball is moving forward at 2 m/s after the collision, what is the change in momentum of the 6-kg ball?

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1. a 4.5-kg ham is thrown at a stationary 15-kg shopping cart. what is the change in momentum of the cart travel if the ham had an initial speed of 2.2 m/s and rebounds off the cart at 0.75 m/s?

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two cars, one twice as heavy as the other, move down a hill at the same speed. compared to the lighter car, the momentum of the heavier car is ______ as much.

1. a moving car has momentum. if it moves twice as fast, its momentum is ______ as much.
2. make two event chains showing what happens when a rolling ball (ball 1) hits a resting ball (ball 2). use the phrases: gains momentum, hits ball 2, is hit by ball 1, loses momentum, rests, rolls, slows down, starts rolling

ball 1
ball 2

1. a car has momentum of 20,000 kg·m/s. what would be the car’s new momentum if

a) its velocity were doubled
b) its velocity were tripled
c) its mass were doubled (by adding more passengers and a greater load)
d) both its velocity were doubled and its mass were doubled

1. which has more momentum, a 1000 kg car moving 1 m/s or a 70 kg person sprinting at 8 m/s?
2. upon hitting a bat, a baseball has a change in momentum of 16.5 kg·m/s. what is the average force applied to the baseball if the time of impact is 0.00200 seconds?

Explanation:

Step1: Define change in momentum

$\Delta p = m(v_f - v_i)$

Step2: Plug in values for 6-kg ball

$m=6\,\text{kg}, v_i=5\,\text{m/s}, v_f=2\,\text{m/s}$
$\Delta p = 6(2 - 5) = 6(-3) = -18\,\text{kg·m/s}$

Step1: Momentum conservation

$m_hv_{hi} + m_cv_{ci} = m_hv_{hf} + m_cv_{cf}$
$v_{ci}=0, v_{cf}=\frac{m_h(v_{hi}-v_{hf})}{m_c}$

Step2: Calculate cart's momentum change

$\Delta p_c = m_cv_{cf} = m_h(v_{hi}-v_{hf})$
$m_h=4.5\,\text{kg}, v_{hi}=2.2\,\text{m/s}, v_{hf}=-0.75\,\text{m/s}$
$\Delta p_c = 4.5(2.2 - (-0.75)) = 4.5×2.95 = 13.275\,\text{kg·m/s}$

Step1: Momentum formula

$p = mv$

Step2: Compare heavy and light cars

Let $m_l=m, m_h=2m, v_l=v_h=v$
$p_h=2mv, p_l=mv \Rightarrow p_h=2p_l$

Step1: Momentum formula

$p = mv$

Step2: Effect of doubled velocity

$v'=2v \Rightarrow p'=m(2v)=2mv=2p$

Answer:

-18 kg·m/s