Question

A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g. what is the velocity acquired by the block?

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Answer 1

Answer:

The velocity acquired by the block is 4.4m/s.

Explanation:

Consider the provided information.

We've been given that a bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g.

With this information, we've been asked to find out the velocity acquired by the block.

Consider the mass of the bullet be $m_1$ gram, mass of the wooden block be $m_2$ gram, the initial velocity of the bullet be $u$ m/s and the final velocity acquired by the block be $v$ m/s.

Mass of the bullet, $m_1 = 10g = \frac{10}{1000}kg = 0.01kg$

Mass of the wooden block, $m_2 = 900g = \frac{900}{1000}kg = 0.9kg$

We can find the final velocity acquired by the block using the moment conservation of law.

According to the law of conservation of linear momentum, for an object or system of objects, the total momentum of the system is always conserved if no external force acts on them.

$m_1 u_1 + m_2v_2 = (m_1 + m_2) \times v$

$\implies v = \dfrac{m_1 u_1 + m_2v_2}{m_1 + m_2} \\ \\ \implies v = \dfrac{0.01 \times 400 + 0.9 \times 0}{0.01 + 0.9} \\ \\ \implies v = \dfrac{4+0}{0.91} \\ \\ \implies v = \dfrac{4}{0.91} \\ \\ \implies \boxed{v = 4.4}$

Hence, the final velocity acquired by the block is 4.4m/s.

Answer 2

Answer:

Given :-

• A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g.

To Find :-

• What is the velocity acquired by the block.

Formula Used :-

$\clubsuit$ Conservation of Momentum Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{(m_1 \times u_1) + (m_2 \times u_2) =\: (m_1 + m_2)v}}}\: \: \: \bigstar$

Solution :-

First, we have to change the units of mass of bullet and wooden block :

☆ In case of mass of bullet (m₁) :

$\implies \sf Mass_{(Bullet)} =\: 10\: g$

$\implies \sf Mass_{(Bullet)} =\: \dfrac{10}{1000}\: kg$

$\implies \sf\bold{\purple{Mass_{(Bullet)} =\: 0.01\: kg}}$

☆ In case of mass of wooden block (m₂) :

$\implies \sf Mass_{(Wooden\: Block)} =\: 900\: g$

$\implies \sf Mass_{(Wooden\: Block)} =\: \dfrac{900}{1000}\: kg$

$\implies \sf\bold{\purple{Mass_{(Wooden\: Block)} =\: 0.9\: kg}}$

Now, we have to find the final velocity acquired by the block :

Given :

• Mass of Bullet (m₁) = 0.01 kg
• Mass of Wooden Block (m₂) = 0.9 kg
• Initial Velocity of Bullet (u₁) = 400 m/s
• Initial Velocity of Wooden Block (u₂) = 0 m/s

According to the question by using the formula we get,

$\footnotesize \implies \bf Momentum_{(Before\: Collision)} =\: Momentum_{(After\: Collision)}$

$\implies \sf\bold{\green{(m_1 \times u_1) + (m_2 \times u_2) =\: (m_1 + m_2)v}}$

By putting those values we get,

$\implies \sf (0.01 \times 400) + (0.9 \times 0) =\: (0.01 + 0.9) \times v$

$\implies \sf (4) + (0) =\: (0.91) \times v$

$\implies \sf 4 + 0 =\: 0.91 \times v$

$\implies \sf 4 =\: 0.91v$

$\implies \sf \dfrac{4}{0.91} =\: v$

$\implies \sf 4.4 =\: v$

$\implies \sf\bold{\red{v =\: 4.4\: m/s}}$

$\therefore$ The final velocity acquired by the block is 4.4 m/s .