A body of mass M moving with velocity v m s- 1 suddenly breaks into two pieces. One part having mass (M/4) remains stationary. The velocity of the other part will be

Question

A body of mass M moving with velocity v m s- 1 suddenly breaks into two pieces. One part having mass (M/4) remains stationary. The velocity of the other part will be (A) v (B) 2v (C) (3 v/4) (D) (4 v/3)

Tardigrade Admin · Accepted Answer

Mv=(M/4)v1+(3/4)Mv2 ⇒ Mv=(3/4)Mv2 (. textAsv1=0.) ⇒ v2=(4 v/3)

Q. A body of mass $M$ moving with velocity $v m s^{- 1}$ suddenly breaks into two pieces. One part having mass $\frac{M}{4}$ remains stationary. The velocity of the other part will be

$v$

$2 v$

$\frac{3 v}{4}$

$\frac{4 v}{3}$

$M v = \frac{M}{4} v_{1} + \frac{3}{4} M v_{2}$
$\Rightarrow M v = \frac{3}{4} M v_{2} (As v_{1} = 0)$
$\Rightarrow v_{2} = \frac{4 v}{3}$

Q. A body of mass M moving with velocity vms−1 suddenly breaks into two pieces. One part having mass 4M​ remains stationary. The velocity of the other part will be

v

2v

43v​

34v​

Mv=4M​v1​+43​Mv2​ ⇒Mv=43​Mv2​(Asv1​=0) ⇒v2​=34v​

Q. A body of mass $M$ moving with velocity $v m s^{- 1}$ suddenly breaks into two pieces. One part having mass $\frac{M}{4}$ remains stationary. The velocity of the other part will be

$v$

$2 v$

$\frac{3 v}{4}$

$\frac{4 v}{3}$

$M v = \frac{M}{4} v_{1} + \frac{3}{4} M v_{2}$
$\Rightarrow M v = \frac{3}{4} M v_{2} (As v_{1} = 0)$
$\Rightarrow v_{2} = \frac{4 v}{3}$