A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the block and that it swings 0.086m above the initial hei…

Question

asked Jan 19, 2021 7.4k views

1 Answer

Tyler Shaddix · Answer 1 · 2021-01-23T09:34:32+0000

Answer:

261.3 m/s

Step-by-step explanation:

Mass of bullet=m=15 g=
(15)/(1000)=0.015 kg

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

(1)/(2)(m+M)^2V^2=(m+M)gh

Using g=
9.8m/s^2

Substitute the values

(1)/(2)(3.015)V^2=3.015* 9.8* 0.086

V^2=(2* 3.015* 9.8* 0.086)/(3.015)

V=√(2* 3.015* 9.8* 0.086)}

V=1.3m/s

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

mv+Mv'=(m+M)V

Using the formula

0.015v+3(0)=3.015(1.3)

0.015v=3.015(1.3)

v^2=(3.015(1.3))/(0.015)=261.3

v=261.3 m/s