QUESTION IMAGE
Question
- during a goal-line stand, a 75-kg fullback moving eastward with a speed of 10 m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. the two players collide and stick together, moving at the same velocity after the collision. determine the the post-collision velocity of the two players.
- calculate the velocity of the rifles recoil after firing.
$m = 4\\ \text{kg}$
$m = 0.010\\ \text{kg}$
$v = 300\\ \text{m/s}$
$v = ?$
- what is the velocity of the \8\ ball after the elastic collision below?
before
$v=2.0\\ \text{m/s}$ $m = 0.25\\ \text{kg}$
$v = 0\\ \text{m/s}$ $m = 0.25\\ \text{kg}$
after
$v = ?\\ \text{m/s}$
- a 6000 kg railroad car moving at 5 m/s collides into a stationary car with a mass of 4000 kg. if they couple together after the collision, what will be their combined velocity immediately after impact?
$v = 5$
$v = 0$
$v = ?\\ \text{m/s}$
Problem 12
Step1: Define variables and direction
Let eastward be positive.
$m_1=75\ \text{kg}$, $v_1=10\ \text{m/s}$; $m_2=100\ \text{kg}$, $v_2=-4\ \text{m/s}$; combined mass $M=m_1+m_2$
Step2: Apply conservation of momentum
Total initial momentum = total final momentum
$$m_1v_1 + m_2v_2 = Mv_f$$
Step3: Solve for final velocity
Substitute values and solve for $v_f$:
$$v_f = \frac{m_1v_1 + m_2v_2}{m_1+m_2} = \frac{(75\times10)+(100\times(-4))}{75+100}$$
$$v_f = \frac{750 - 400}{175} = \frac{350}{175}$$
Step1: Define variables (momentum conservation)
Initial total momentum = 0 (system at rest). $M=4\ \text{kg}$, $m=0.010\ \text{kg}$, $V=300\ \text{m/s}$
Step2: Set up momentum equation
$$0 = Mv + mV$$
Step3: Solve for recoil velocity $v$
$$v = -\frac{mV}{M} = -\frac{0.010\times300}{4}$$
Step1: Recall elastic collision rules (equal mass)
For elastic collision of two identical masses ($m_1=m_2=0.25\ \text{kg}$), the first mass stops, second takes its velocity.
Step2: Verify with momentum/KE conservation
Initial momentum: $m\times2.0 + m\times0 = 2m$; Final momentum: $m\times0 + m\times v_f = mv_f$, so $v_f=2.0\ \text{m/s}$
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$2\ \text{m/s (eastward)}$
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