First slide

The steps in collision

Question

A particle of mass 1 kg is thrown vertically upward with speed $100 {ms}^{- 1}$ . After 5 s, it explodes into two
parts. One part of mass 400 g emerges with speed $25 {ms}^{- 1}$ in downward direction, what is the velocity of other part just after explosion? (Take, $g = 10 {ms}^{- 2}$ )

Moderate

A

$100 {ms}^{- 1}$ upward
B

$600 {ms}^{- 1}$ upward
C

$100 {ms}^{- 1}$ downward
D

$300 {ms}^{- 1}$ upward

Solution

Velocity of particle after 5 s,
$v = u = g t = 100 - 10 \times 5$
$= 100 - 50 = 50 {ms}^{- 1}$ (upwards)
Conservation of linear momentum gives
$Mv = m_{1} v_{1} + m_{2} v_{2}$ …(i)
Taking upward direction positive,
$v_{1} = - 25 {ms}^{- 1}, v = 50 {ms}^{- 1}$
$M = 1 kg, m_{1} = 400 g = 0.4 kg$
$m_{2} = M - m_{1} = 1 - 0.4 = 0.6 kg$
From Eq. (i), we get
$1 \times 50 = 0.4 \times (- 25) + 0.6 v_{2} or v_{2} = 100 {ms}^{- 1} (upwards)$

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