Question

A bullet of mass 0.05 kg moving with velocity 400 m/s gets embedded after collision with a block ofmass 3.95 kg placed on a smooth horizontalplatform as shown in the figure. If height of theplatform from the ground is 20 m and block strikesthe ground at B, then the value of AB is(Take g = 10 m/s2) solution plzzzzz ​

Answer 1

Given:

A bullet of mass 0.05 kg moving with velocity 400 m/s gets embedded after collision with a block of mass 3.95 kg placed on a smooth horizontal platform. The platform is at height of 20 m

To find:

Value of AB (Range of the projectile)

Calculation:

Applying Conservation of Momentum :

$\therefore \: (m1)(v1) = (m1 + m2)v2$

$= > (0.05 \times 400) = (0.05 + 3.95) \times v2$

$= > (0.05 \times 400) = 4\times v2$

$= > v2 = 5 \: m {s}^{ - 1}$

Now , the block-bullet combination shall follow the height to ground projectile as shown in the attached diagram.

Range of the projectile is equal to AB.

$AB = velocity \times time$

$= > AB = 5 \times \sqrt{ \dfrac{2h}{g} }$

$= > AB = 5 \times \sqrt{ \dfrac{2 \times 20}{10} }$

$= > AB = 5 \times 2$

$= > AB = 10 \: m$

So final answer:

AB is equal to 10 m