Question

A 10 gram bullet travelling at 200 m per second strikes and remains embedded in a 2kg target which is originally at rest but free to move at what speed does the target move of

Answer 1

Mass of the bullet (m) = 0.01 kg

Initial velocity of bullet (u) = 200 m/s

Let the common velocity after the collision be v.

Mass of the target (M) = 2 kg

Initial momentum = M × 0 + 0.01 × 200 = 2 kg m/s

Final momentum = (M + m) × v

Since, the forces between block and bullet are internal forces. Hence, the momentum of the system will be conserved.

Initial momentum = Final momentum

(2 + 0.01) × v = 2

2.01 × v = 2

v = 0.99 m/s

Hence, the common velocity of target and bullet is 0.99 m/s after the collision.

Answer 2

Given,

m₁ = 10 g = 10/100 kg

m₂ = 2 kg

u₁ = 200 m/s

u₂ = 0 m/s

                                                       

Here we use formula

m₁u₁ + m₂u₂ = (m₁+m₂) v

                                                 

By using this formula we get,

$\dfrac{10}{1000}\times200+2\times0=\bigg(\dfrac{10}{1000}+2\bigg)\:v\\\\\\\implies2+2\times0=\bigg(\dfrac{10+2000}{1000}\bigg)\:v\\\\\\\implies2=\bigg(\dfrac{2010}{1000}\bigg)\:v\\\\\\\implies2=\dfrac{201}{100}\times v\\\\\\\implies2\times100=201\times v\\\\\\\implies200=201\times v\\\\\\\implies\dfrac{200}{201}=v\\\\\\\therefore\:v=\boxed{\bold{0.99\:m/s.}}$

So the target will move at the speed of 0.99 m/s.