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evablogger [386]

3 years ago

6

A bullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sittin

g at rest on a nearly frictionless surface.(a) What is the speed of the block after the bullet embeds itself in the block?(b) Calculate the total translational kinetic energy before and after the collision.

1 answer:

rodikova [14]3 years ago

6 0

Answer:

(a) vf = 0.98 m/s

(b) K₁ = 714.85 J : Total translational kinetic energy before the collision.

K₂= 2.41 J : Total translational kinetic energy after the collision.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁ = 0.017 kg : mass of the bullet

m₂ = 5 kg : mass of the block

v₀₁ = 290 m/s : initial velocity of the bullet

v₀₂ = 0 : initial velocity of the block₂

(a) Speed of the block after the bullet embeds itself in the block

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf

vf: final velocity of the block

( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf

4.93+ 0 = ( 5.017 )*vf

vf = 4.93 / 5.017

vf = 0.98 m/s

b) Total translational kinetic energy before (K₁) and after the collision(K₂).

K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)

K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J

K₂= 1/2(m₁+ m₂)*vf²

K₂= 1/2(0.017+ 5)*(0.98)² = 2.41 J

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