Question

A mass $m$ moving horizontally (along the $x-axis)$ with velocity $v$ collides and sticks to a mass of $3m$ moving vertically upward (along the $y$ -axis) with velocity $2\,v$. The final velocity of the combination is

• A. $\frac{3}{4} v \widehat{i} + \frac{1}{4} v \widehat{j}$
• B. $\frac{1}{4} v \widehat{i} + \frac{3}{2} v \widehat{j}$
• C. $\frac{1}{3} v \widehat{i} + \frac{2}{3} v \widehat{j}$
• D. $\frac{2}{3} v \widehat{i} + \frac{1}{3} v \widehat{j}$

Answer 1

The Correct Option is B

To find the final velocity of the combined masses after the collision, we can use the principle of conservation of momentum. Since the collision is perfectly inelastic, the masses stick together after the collision, and we treat each direction separately.

Step 1: Initial Momentum Calculation

1. Initial momentum along the x-axis (p_{x_{\text{initial}}}):
   • Only mass m has velocity along the x-axis.
   • Momentum of mass m: p_{x_{\text{m}}} = mv
   • Total initial x-momentum: mv
2. Initial momentum along the y-axis (p_{y_{\text{initial}}}):
   • Only mass 3m has velocity along the y-axis.
   • Momentum of mass 3m: p_{y_{3m}} = 3m \cdot 2v = 6mv
   • Total initial y-momentum: 6mv

Step 2: Total Mass and Final Momentum Calculation

The combined mass after collision is m + 3m = 4m.

1. Using the conservation of momentum:
   • Final velocity along x-axis (v_x):
   • 4m \cdot v_x = mv
   • Solving: v_x = \frac{mv}{4m} = \frac{v}{4}
2. Using the conservation of momentum:
   • Final velocity along y-axis (v_y):
   • 4m \cdot v_y = 6mv
   • Solving: v_y = \frac{6mv}{4m} = \frac{3v}{2}

Step 3: Final Velocity

The final velocity \vec{v}_{\text{final}} of the system is a combination of the x and y components:

\vec{v}_{\text{final}} = v_x \widehat{i} + v_y \widehat{j} = \frac{1}{4}v \widehat{i} + \frac{3}{2}v \widehat{j}

Thus, the correct answer is: \frac{1}{4} v \widehat{i} + \frac{3}{2} v \widehat{j}