A block of mass 1 kg moving with a speed of $4 m s^{- 1}$ , collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The loss in the K.E of the system

8 J

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4 \times 10^{- 7}

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4 J

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0 J

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Solution

The correct option is C $4 J$
Conserving momentum
$4 \times 1 = 2 \times v$
$\Rightarrow v = 2 m / s$

$Δ K . E . = \frac{1}{2} \times 1 \times (4)^{2} - \frac{1}{2} \times 2 \times (2)$

$= 4 J$

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A block of mass 1 kg moving with a speed of4ms−1, collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The loss in the K.E of the system

The correct option is C 4JConserving momentum4×1=2×v⇒v=2m/sΔK.E.=12×1×(4)2−12×2×(2)=4J

A block of mass 1 kg moving with a speed of $4 m s^{- 1}$ , collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The loss in the K.E of the system

The correct option is C $4 J$
Conserving momentum
$4 \times 1 = 2 \times v$
$\Rightarrow v = 2 m / s$

$Δ K . E . = \frac{1}{2} \times 1 \times (4)^{2} - \frac{1}{2} \times 2 \times (2)$

$= 4 J$