Question

A bullet of mass 10 g moving with 100 m/s is embedded in a block of 1 kg which is initially at rest.The final velocity of the system will be

Answer 1

mass of bullet =m1=10g=10/1000=0.001kg

initial velocity of bullet=v1=100m/s

mass of block=m2=1kg

initial velocity of block=v2=0

we have to find final velocity of system

we will assume this system as an elastic system

we will use formulas

m1v1+m2v2=m1v1/+m2v2/.............(1)

v1+v1/=v2+v2/........................(2)

v1/ and v2/ are final velocities of system after collision

put values in equation (1)

m1v1+m2v2=m1v1/+m2v2/

(0.001)(100)+(1)(0)=(0.001)(v1/)+1(v2/)

1=0.001v1/+v2/...........(3)

now take equation (2) and put values in that

v1+v1/=v2+v2/

100+v1/=0+v2/

100=v2/-v1/.................(4)

now multiplying equation (3) by negative and  then add  intoequation 4 we get

1=0.001v1/+v2/

-1= -(0.001v1/+v2/)

-1= -0.001v1/-v2/.............(5)

adding up these two equation (4) and (5) we get

100-1=v2/-v2/-v1/+-0.001v1/

99=  - 1.001v1/

v1/= -98.90m/s

negative sign shows that velocity will in the opposite site

now put  value of v1. in (4) to get value of v2/

100=v2/-v1/

100=v2/-(-98.9)

100-98.9=v2/

v2/=0.2m/s

so the final velocity will be

v1/= -98.9m/s

v2/=0.2m/s