The Question is from neet-physics electromagnetic-waves , and it was created by the author VENKATESWARARAO

Question

A 0.5 kg ball moving at 4 m/s collides head-on with a stationary 1 kg ball. If the collision is perfectly elastic, what is the velocity of the 0.5 kg ball after the collision?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Conservation of momentum and kinetic energy in elastic collisions
 **Related Topics**: Collision theory, linear momentum
 **Prerequisites**: Understanding of elastic collisions, knowledge of momentum and kinetic energy formulas.
## SOLUTION PATHWAY
**Given Information**:
  $m_1 = 0.5 kg$ (mass of first ball)
  $u_1 = 4 m/s$ (initial velocity of first ball)
  $m_2 = 1 kg$ (mass of second ball)
  $u_2 = 0 m/s$ (initial velocity of second ball)
  Collision is perfectly elastic.
**Method & Steps**:
 1. Apply the conservation of momentum principle.
  Reasoning: In an elastic collision, the total momentum before the collision equals the total momentum after the collision.
  Formula/Rule: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$, where $v_1$ and $v_2$ are the final velocities of the balls.
 2. Apply the conservation of kinetic energy principle.
  Working:  $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$
  Substitute the given values into the momentum equation: $0.5(4) + 1(0) = 0.5v_1 + 1v_2  => 2 = 0.5v_1 + v_2$.
  Substitute the given values into the kinetic energy equation: $\frac{1}{2}(0.5)(4)^2 + 0 = \frac{1}{2}(0.5)v_1^2 + \frac{1}{2}(1)v_2^2 => 4 = 0.25v_1^2 + 0.5v_2^2$.
  Solve the system of two equations with two unknowns ($v_1$ and $v_2$).  From the momentum equation: $v_2 = 2 - 0.5v_1$. Substitute this into the kinetic energy equation and solve the quadratic equation for $v_1$.
 3. Solve for $v_1$ (velocity of the 0.5 kg ball after collision).
  Result: Solving the quadratic equation yields two solutions for $v_1$: $v_1 = 4$ (this is the initial velocity, which is not the solution after collision) and $v_1 = -4/3 ≈ -1.33 m/s$.
  Verification: Substitute $v_1 = -4/3$ back into the momentum equation to find $v_2 = 8/3$.  Check if these values satisfy the kinetic energy equation. They do.
 ## OPTION ANALYSIS
A: :x: This is the incorrect sign. The 0.5 kg ball reverses direction after the collision.
B: ✓: This is the correct velocity of the 0.5 kg ball after the collision. The negative sign indicates that it moves in the opposite direction.
C: :x: Incorrect magnitude and sign.
D: :x: Incorrect magnitude and sign.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying the conservation laws: Ensure both momentum and kinetic energy are conserved in an elastic collision.
  2.Mathematical errors: Carefully solve the system of equations. Check your calculations.
 **Quick Tips**:
  1.For head-on elastic collisions, use the equations for velocity after collision directly:  $v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}$ and $v_2 = \frac{2m_1u_1 + (m_2 - m_1)u_2}{m_1 + m_2}$.
  2.Remember that the negative sign indicates a change in direction.

VENKATESWARARAO · Answer

1.33 m/s

VENKATESWARARAO · Answer

The correct Option is "-1.33 m/s"

VENKATESWARARAO · Answer

2.67 m/s

VENKATESWARARAO · Answer

-2.67 m/s

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