Question

Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marble A was initially moving at a velocity of 0.7 m/s, but after the collision it has a velocity of –0.2 m/s. What is the resulting velocity of marble B after the collision?

Answer 1

Answer:the resulting velocity of marble B after collision is $1.8 \frac{m}{s}$

Explanation:Consider marble A and marble B as a single system

Now apply law of conservation of linear momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b$

where $m_a=0.06 kg, u_a=0.7 \frac{m}{s}, v_a=-0.2 \frac{m}{s}$

$m_b=0.03 kg, u_b=0.0 \frac{m}{s}, v_b=? \frac{m}{s}$

Therefore $0.06\times 0.7+0.03\times 0=(0.06\times -0.2)+0.03v_b$

$v_b=1.8 \frac{m}{s}$

Thus the resulting velocity of marble B after collision is 1.8 m/s

Answer 2

We can solve this using the Law of Conservation of Momentum. If both marbles are in our system, the initial momentum should equal the final momentum.

The initial momentum can be solved for as so:

$m_(a)$ * $v_(a)$ + $m_{b} * v_{b}$ = $p_{o}$
(0.06)(0.7) + (0.03)(0) = 0.042 [kg * m/s]

So if the system has an initial momentum of 0.042, it should have the same final momentum.

$m_{a} * v_{a,f} + m_{b} * v_{b,f} = 0.042$
(0.06)(-0.2) + (0.03)($v_{b,f}$) = 0.042
(0.03)($v_{b,f}$) = 0.54
($v_{b,f}$) = 18 [m/s]