Question

A bullet of mass m=11 gr is fired into a block of mass M=1 kg initially at rest at the edge of a frictionless table of height h=1.10 m. The bullet remains in the block and the block lands a distance d=0.67 m from the bottom of the table.

a)Determine the initial velocity of the bullet.

b) Determine the loss of kinetic Energy during the collision.

Answer 1

Part A.

First calculate the initial speed of projectile motion when Block+Bullet starts falling

Since there is no initial vertical velocity for Bullet + Block, So Using 2nd kinematic equation:

h = V0y*t + (1/2)*a*t^2

h = vertical distance traveled = -1.10 m

V0y = 0 m/sec, Since initial velocity is only in horizontal direction

a = Vertical acceleration = -g = -9.81 m/sec^2

So,

-1.10 = 0 + (1/2)*(-9.81)*t^2

t = sqrt (2*1.10/9.81) = 0.47356 sec

Now range in projectile motion is given by:

R = V0x*t

R = horizontal distance traveled by block + bullet = d = 0.67 m

V0x = Initial speed of block + bullet = R/t

V0x = 0.67/0.47356 = 1.4148 m/sec

Now Using momentum conservation on block + bullet before and after collision

Pi = Pf

mb*vb + Mb*Vb = (mb + Mb)*V0x

mb = mass of bullet = 11.0 gm = 0.011 kg

Mb = mass of block = 1 kg

Vb = initial speed of block = 0 m/sec

vb = initial speed of bullet = ?

So,

vb = (mb + Mb)*V0x/mb

vb = (0.011 + 1)*1.4148/0.011 = 130.033 m/s

vb = Initial speed of bullet = 130.0 m/sec

Part B.

Now loss of kinetic energy during the collision will be:

dK = KEi - KEf

dK = (1/2)*mb*vb^2 + (1/2)*Mb*Vb^2 - (1/2)*(mb + Mb)*V0x^2

Using known values:

dK = (1/2)*0.011*130.033^2 + (1/2)*1*0^2 - (1/2)*(0.011 + 1)*1.4148^2 = 91.985 J

dK = 92.0 J = Loss in kinetic energy

Let me know if you've any query.