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Grade 11Mechanics

A wooden block of mass 10 kg is suspended with a long string. A bullet of mass 50 g strikes the block and gets embedded in it. As a result, block swings to a maximum height of 10 cm. Find the initial velocity of the bullet.

archi

10 Years agoGrade 11

1 Answer

Amogh Gajare

Approved Tutor Answer10 Years ago

Hello!

Principles used in the problem –

Principle of conservation of momentum & Work Energy theorem.

Given information -

Mass of bullet = 50 g = 0.05 kg

Mass of block = 10 kg

Solution -

To find Velocity ‘v’ of bullet,

By principle of conservation of linear momentum,

$m_1*v_1 = m_2*v_2$

Initially, mass in block was just 10 kg.But after bullet got embeeded into it, bullet’s weight got added.

So,

$0.05*v_1 = (10 + 0.05)*v_2$ …..................Equation 1

By work energy theorem,

Work done = change in kinetic energy

$- mgh = K_f - K_i.............$ as reference level is taken above, it gives a negative sign to work)

(K $\rightarrow$ Kinetic energy)

$- mgh = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2$
As block was initially at rest, V₁ = 0

So

$- mgh = - \frac{1}{2}mv_2^2$

Substituting values and calculating we get,

$v_2 = \sqrt2$ …............................................................Equation 2

From Equation 1 & 2 we get,

$0.05*v_1 = 10.05*\sqrt2$

On calculation,

$v_1 = 201\sqrt2$ m/s

If any further doubts, feel free to email me at -amoghsgajare @gmail.com

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