Question

A bullet of mass 10 g travelling at a speed of 500 ms–1 strikes a block of mass 2 kg, which is suspended by a string of length 5 m. the centre of gravity of the block is found to rise a vertical distance of 0.2 m. what is the speed of the bullet after it emerges from the block?

Answer 1

Answer:

                                 Speed of the bullet is 106m/s.

Explanation:

Given Data :

Mass of bullet $(m)= 10gm =0.01kg$

Initial speed of bullet $(v)=500m/s$

Mass of Block $(M)=2kg$

length of the string $(l)=5m$

Height/vertical distance of block $(h)=0.1m$

Velocity can be give as;

          Kinetic Energy of Block=Potential Energy of Block

$\frac{1}{2}MV^{2}=mgh\\V=\sqrt{2gh}\\V=\sqrt{2*9.8*0.2}\\V=1.97m/s$

Suppose, speed of bullet is$v^{'}$,speed of bullet can be found by principle of linear momentum:

          Initial momentum=Final Momentum

$P_{bullet}=P_{block}+P_{bullet}\\   m*v+M*0=M*V+mv^{'}\\0.01*500=2*1.97+0.01v^{'}\\ 5=3.94+0.01v^{'}\\\frac{ 5-3.94}{0.01} =v^{'}\\v^{'}=106m/s$

Answer 2

mass of bullet ......0.01

that is 1000 g ---> I kg

Ig-----> I/1000

10g ----> I/1000x10=10^-2

height (h)=0.2