Question

bullet of mass m=14 gr is fired into a block of mass M=2 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=0.66 m from the bottom of the table.

a)Determine the initial velocity of the bullet.
v_i=  m/s
b) Determine the loss of kinetic Energy during the collision.
ΔK =  J

Answer 1

a) as the block with bullet embedded in it fall 0.66 m from the foot of table. and height of the table is 1.10 m. then using vertical component of second equation of motion,

h = 0 + (0.5)*g*t²

t = √(2*h/g)

using known values,

t = √(2.2/9.81)

t = 0.47 seconds

Now the speed of block after hitting by the bullet is,

Vf = d/t

Vf = 0.66/0.47

Vf = 1.39 m/s

Now using conservation of momentum,

initial momentum = Final momentum

m*vi + M*Vi = (m + M) * Vf

using the known values,

0.014*vi + 0 = 2.014 * 1.39

solving for vi,

vi = 200.5 m/s

b) the initial kinetic energy of the system,

Ki = 0.5 * [ m * vi² + M * Vi²]

Ki = 0.5 * [0.014 * 200.5² + 0]

Ki = 281.38 J

Now, the final Kinetic energy,

Kf = 0.5 * (m + M) * Vf²

Kf = 0.5 * 2.014 * 1.39²

Kf = 1.95 J

Thus the loss in kinetic energy during collision is,

∆k = Ki - Kf

∆k = 281.38 - 1.95

∆k = 279.37 J

Final Answers:-

a) 200.5 m/s

b) 279.37 J