Question

A bullet of mass m=20 gr is fired into a block of mass M=5 kg initially at rest at the edge of a frictionless table of height h=0.90 m. The bullet remains in the block and the block lands a distance d=0.68 m from the bottom of the table. Picture1 a)Determine the initial velocity of the bullet. vi= m/s b) Determine the loss of kinetic Energy during the collision. ΔK = J.

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 = -0.90 m

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

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

So,

-0.90 = 0 + (1/2)*(-9.81)*t^2

t = sqrt (2*0.90/9.81) = 0.42835 sec

Now range in projectile motion is given by:

R = V0x*t

R = horizontal distance traveled by block + bullet = d = 0.68 m

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

V0x = 0.68/0.42835 = 1.5875 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 = 20.0 gm = 0.020 kg

Mb = mass of block = 5 kg

Vb = initial speed of block = 0 m/sec

vb = initial speed of bullet = ?

So,

vb = (mb + Mb)*V0x/mb

vb = (0.020 + 5)*1.5875/0.020 = 398.4625 m/s

vb = Initial speed of bullet = 398.5 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)*0.020*398.4625^2 + (1/2)*5*0^2 - (1/2)*(0.020 + 5)*1.5875^2

dK = 1581.4 J = Loss in kinetic energy (since we need to calculate loss in KE, So use positive value as your final answer)

Let me know if you've any query.