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?
Answer:
The resulting velocity of marble B after the collision is 0.96 m/s.
Explanation:
Mass of marble A ,\(m_1= 0.08 kg\)
Initial velocity of the marble A,\(u_1 = 0.5 m/s\)
Mass of marble B,\(m_2 = 0.05 kg\)
Initial velocity of the marble B,\(u_2 = 0 m/s\)
final velocity of the marble A,\(v_1 = -0.1 m/s\)
final velocity of the marble B,\(v_2 = ?\)
Applying law of conservation of mass:
\(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)
\(0.08 kg\times 0.5 m/s+0.05kg\times 0 m/s=0.08 kg\times (-0.1 m/s)+0.05kg \times v_2\)
\(v_2=0.96 /s\)
The resulting velocity of marble B after the collision is 0.96 m/s.
Answer:
the other ball will move with speed 0.96 m/s
Explanation:
Here in collision type of questions we can use momentum conservation because there is no force on the system
So here we will have
\(m_a v_{1i} + m_b v_{2i} = m_a v_{1f} + m_b v_{2f}\)
so here we have
\(0.08(0.5) + 0.05(0) = 0.08(-0.1) + 0.05 v\)
\(0.048 = 0.05 v\)
\(v = 0.96 m/s\)
So the light ball will move with speed 0.96 m/s