A $10 g$ bullet traveling at $200 m / s$ strikes and remains embedded in a $2 k g$ target which is originally at rest but free to move. At what speed does the target move off?

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Solution

Given,
$m_{1} = 10 g = 0.01 k g$
$v_{1} = 200 m / s$
$m_{2} = 2 k g$
$v_{2} = ?$
From the conservation of linear momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$0.01 \times 200 = 2 v_{2}$
$v_{2} = \frac{0.01 \times 200}{2} = 1 m / s$

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A 10g bullet traveling at 200m/s strikes and remains embedded in a 2kg target which is originally at rest but free to move. At what speed does the target move off?

Given,m1=10g=0.01kgv1=200m/sm2=2kgv2=?From the conservation of linear momentum,m1v1=m2v20.01×200=2v2v2=0.01×2002=1m/s

A $10 g$ bullet traveling at $200 m / s$ strikes and remains embedded in a $2 k g$ target which is originally at rest but free to move. At what speed does the target move off?

Given,
$m_{1} = 10 g = 0.01 k g$
$v_{1} = 200 m / s$
$m_{2} = 2 k g$
$v_{2} = ?$
From the conservation of linear momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$0.01 \times 200 = 2 v_{2}$
$v_{2} = \frac{0.01 \times 200}{2} = 1 m / s$