Question

Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of –0.1 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.

Answer 1

Explanation:

Given that,

Mass of marble A, $m_A=0.08\ kg$

Mass of marble B, $m_B=0.05\ kg$

Marble B is still, $u_B=0$

Initial speed of marble A, $u_A=0.5\ m/s$

Final speed of marble A after the collision, $v_A=-0.1\ m/s$

We need to find the resulting velocity of marble B after the collision. It can be calculated using conservation of momentum as :

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B$

$0.08\times 0.5+0.05\times 0=0.08\times (-0.1)+0.05v_B$

$v_B=\dfrac{0.032}{0.05}=0.64\ m/s$

So, the resulting velocity of marble B after the collision is 0.64 m/s. Hence, this is the required solution.

Answer 2

(M1xVi1)+(M2xVi2)=(M1xVf1)+(M2xVf2)
(.08x.5)+(.05x0)=(.08x-.1)+(.05xVf2)

Marble B's resulting velocity is .64