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A body of mass $3 \mathrm{~kg}$ is moving with a velocity of $8 \mathrm{~ms}^{-1}$ collides head on with another body of mass $1 \mathrm{~kg}$, moving in the opposite direction with a velocity of $4 \mathrm{~ms}^{-1}$. After the collision, if the two bodies stick together and move, they move with a common velocity
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Verified Answer
The correct answer is:
$5 \mathrm{~ms}^{-1}$
Given, mass, $m_1=3 \mathrm{~kg}$
Velocity, $v_1=8 \mathrm{~ms}^{-1}$
Mass of another body, $m_2=1 \mathrm{~kg}$
Velocity of another body, $v_2=4 \mathrm{~ms}^{-1}$
After collision, both bodies stick together and move with a common velocity, applying
conservation of momentum,
$m_1 v_1-m_2 v_2=\left(m_1+m_2\right) v$
where, v is the common velocity.
$\begin{aligned} & (3 \times 8)-(1 \times 4)=(3+1) \times v \\ \Rightarrow \quad v & =\frac{20}{4} \\ & =5 \mathrm{~ms}^{-1}\end{aligned}$
Velocity, $v_1=8 \mathrm{~ms}^{-1}$
Mass of another body, $m_2=1 \mathrm{~kg}$
Velocity of another body, $v_2=4 \mathrm{~ms}^{-1}$
After collision, both bodies stick together and move with a common velocity, applying
conservation of momentum,
$m_1 v_1-m_2 v_2=\left(m_1+m_2\right) v$
where, v is the common velocity.
$\begin{aligned} & (3 \times 8)-(1 \times 4)=(3+1) \times v \\ \Rightarrow \quad v & =\frac{20}{4} \\ & =5 \mathrm{~ms}^{-1}\end{aligned}$
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