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 height. Calculate the velocity of the bullet when it strikes the wood.

261.3 m/s

Explanation:

Mass of bullet=m=15 g=

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

Using g=

Substitute the values

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

Using the formula

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

conservation of mass

mass before something happens = mass afterwards