Question

A mass M moving with velocity V along X-axis collides and sticks to another mass 2M which is moving along Y-axis with velocity 3V. The velocity of the combination, after the collision is

• A. \(V\hat{i}+2V\hat{j}\)
• B. \(\frac{V}{3}\hat{i}+2V\hat{j}\)
• C. \(V\hat{i}+3V\hat{j}\)
• D. \(2V\hat{i}+4V\hat{j}\)

Hint:

Conserve momentum separately along X and Y; the total mass is 3M.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The masses stick together, so the collision is perfectly inelastic. Linear momentum is conserved in each direction.

Step 2: Momentum before the collision:
Along X: \(M\cdot V=MV\). Along Y: \(2M\cdot3V=6MV\).

Step 3: Momentum after the collision:
The combined mass is \(3M\) with velocity \((v_x,v_y)\). \(3Mv_x=MV\) gives \(v_x=\dfrac V3\). \(3Mv_y=6MV\) gives \(v_y=2V\).

Step 4: Result:
\(\vec v=\dfrac V3\hat i+2V\hat j\). Option B.

Step 5: Why the other options are wrong.
Options A, C and D divide by 3 only partially or not at all, ignoring that the combined mass is \(3M\).

Final Answer:
The velocity is (V/3) i + 2V j. \[ \boxed{\text{(B) }\dfrac V3\hat i+2V\hat j} \]