Question

consider a 10 g mass traveling at 10 m/s impacting and sticking to a 990 g mass that is initially at rest. what is the speed of the combined mass after the collision (in m/s)?
a 0.01
b 0.1
c 10
d 100

Explanation:

Step1: Apply conservation of momentum

Momentum before collision = momentum after collision.
Momentum \(p = mv\).
Before collision: \(p_{initial}=m_1v_1 + m_2v_2\), where \(m_1 = 10\space g=0.01\space kg\), \(v_1 = 10\space m/s\), \(m_2 = 990\space g = 0.99\space kg\), \(v_2=0\space m/s\).
So \(p_{initial}=0.01\times10+0.99\times0 = 0.1\space kg\cdot m/s\).
After collision: \(p_{final}=(m_1 + m_2)v\), where \(m_1 + m_2=0.01 + 0.99=1\space kg\).

Step2: Solve for \(v\)

Since \(p_{initial}=p_{final}\), we have \(0.1=(1)v\).

Answer:

B. 0.1