Question

12. (20 points) A 5 gram bullet is fired at a wood block that is supported at rest on a smooth incline plane (no friction) at an angle of theta above the horizontal. The bullet is fired with a speed of 500 m/s, but then impacts and passes straight through the wood block, coming out the other side with a speed of 400 m/s. As a result of the collision, the wood block slides up the ramp 5 cm from its starting point. Find the mass of the wood block in terms of the mass of the bullet (mg), initial and final velocity of the bullet (V. and Vf), gravity (g), the angle theta , and the final height (y) of the wood block vertically above its lowest point (define this as y.). 5 cm 400 m/s ---- Yi 500 m/s y = 0 m

Answer 1

12.

Consider the motion of the block just after bullet emerge out of it :

d = distance traveled by the block parallel to incline surface = 5 cm = 0.05 m

$\theta$ = Angle of incline surface

h = height gained by the block = Yf - Yo

we know that , Sin$\theta$ = h/d

Yf - Yo = d Sin$\theta$

d = (Yf - Yo )/Sin$\theta$ Eq-1

vf = final velocity of the block after collision

M = mass of the wood block = ?

Using conservation of energy

Potential energy gained by block = Kinetic energy

Mgh = (0.5) M vf2

vf2 = 2gh

vf = sqrt(2 g (Yf - Yo) ) Eq-2

mB = mass of the bullet = 5 g = 0.005 kg

Vo = initial velocity of the bullet before collision = 500 m/s

Vf = final velocity of the bullet after collision = 400 m/s

vi = initial velocity of the block before collision = 0 m/s

Using conservation of momentum

mB Vo + M vi = mB Vf + M vf

mB Vo + M (0) = mB Vf + M ( sqrt(2 g (Yf - Yo) ) )

M =  mB (Vo -  Vf )/ sqrt(2 g (Yf - Yo) )