Question

i GOT DISCONNECTED WITH A TUTOR THAT WAS EXPLAING, NEED HELPIna 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

Given that,

The mass of marble A, m₁=0.08 kg

The mass of marble B, m₂=0.05 kg

The initial velocity of marble A, u₁=0.5 m/s

As the marble B was at rest, the initial velocity of marble B is u₂=0 m/s

The final velocity of marble A, v₁=-0.1 m/s

Let the final velocity of marble B be v₂.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total after the collision.

Therefore

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

On rearranging the above equation,

$v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

On substituting the known values in the above equation,

$\begin{gathered} v_2=\frac{0.08\times0.5+0.05\times0-0.08\times-0.1}{0.05} \\ =\frac{0.048}{0.05} \\ =0.96\text{ m/s} \end{gathered}$

Therefore the

Answer 2

Answer:

See below

Explanation:

Use conservation of momentum (rather than Kinetic Energy ) for collision problems:

Momentum A  = mv = .08 * .5 = .04  km m/s

Momentum B = mv  = 0

    Total Momentum = .04 + 0 = .04  kg m/s

After collision the sum must be the same

A = mv = .08 ( -.1 ) = - .008 kg m/s

B = m vb   = .05 vb

    .04   =  - .008   +  .05 vb

      vb =  + .96 m/s      in the same direction as original A direction