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?

 Solution:-Given that

m1=10g=0.01 kg

v1=200 m/s

m2=(2+m1)kg=(2+0.01) kg=2.01 kg

​v2=? 

From the conservation of linear momentum,

m1v1=m2v2

​0.01×200=2.01 ×v2

v2=0.01×200/2.01 

v2=200/2.01=0.99 m/s

The target speed is 0.99 m/s