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

Question

A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to the mass of 3m moving vertically upward (along the y-axis) with velocity 2v . The final velocity of the combination is (A) (1/4)v hati+(3/2)v hatj (B) (1/3)v hati+(2/3)v hatj (C) (2/3)v hati+(1/3)v hatj (D) (3/2)v hati+(1/4)v hatj

Tardigrade Admin · Accepted Answer

As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.
Solution

∴ m v \hat{i} + 3 m (2 v) \hat{j} = (4 m) v \to

v \to = \frac{v}{4} \hat{i} + \frac{6}{4} v \hat{j} = \frac{v}{4} \hat{i} + \frac{3}{2} v \hat{j}

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

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.

$∴ m v \hat{i} + 3 m (2 v) \hat{j} = (4 m) v \to$
$v \to = \frac{v}{4} \hat{i} + \frac{6}{4} v \hat{j} = \frac{v}{4} \hat{i} + \frac{3}{2} v \hat{j}$

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

41​vi^+23​vj^​

31​vi^+32​vj^​

32​vi^+31​vj^​

23​vi^+41​vj^​