A 5.00-g bullet moving with an initial speed of V1 = 420 m/s is fired into and p

Question

A 5.00-g bullet moving with an initial speed of V1 = 420 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 920 N/m. The block moves d = 5.80 cm to the right after impact before being brought to rest by the spring. (a) Find the speed at which the bullet emerges from the block. m/s (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet?block system during the collision. J

Explanation / Answer

a)

let m = 5 grams = 0.005 kg
M = 1 kg

let v is the speed of Block after the collsion.

Now Apply Enrgy conservation after the collsion,

0.5*k*d^2 = 0.5*M*v^2

v = d*sqrt(k/M)

= 0.058*sqrt(k/1)

k value is not present in the question.
sustitute k value in the above equation and get v.

Now Apply momentum conservation

Initial momentum = final momentum

m*vi = M*v + m*vf

vf = (m*vi - M*v)/m

= (0.005*420 - 1*v)/0.005

= .... m/s

b) thermal enrgy created = Ki - Kf

= 0.5*m*vi^2 - (0.5*m*vf^2 + 0.5*M*v^2)

substitute vf and v values in the above equation and get the required value.

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