Question

A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling. The block swings in an arc, rising 3 mm above its lowest position. The velocity of the bullet was:

Answer 1

Answer:

Velocity of the bullet was 800 m/s.

Explanation:

It is given that,

Mass of the block, m₁ = 10 kg

Mass of the bullet, m₂ = 3 g

Height reached by the block, h = 3 mm

Let v is the velocity of the wooden block. Using the conservation of energy to find it as :

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 3\times 10^{-3}}$

v = 0.24 m/s

Let v' is the velocity of the bullet. The momentum of the system remains conserved. It can be calculated as :

$m_1v=m_2v'$

$v'=\dfrac{m_1v}{m_2}$

$v'=\dfrac{10\times 0.24}{3\times 10^{-3}}$

v' = 800 m/s

So, the velocity of the bullet was 800 m/s. Hence, this is the required solution.

Answer 2

The velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.

What is conservation of momentum?

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling.

The kinetic energy will be equal to the gravitation potential energy for this case. Thus,

$\dfrac{1}{2}mv^2=mgh\\v^2=2gh$

As the maximum height is 3mm or 0.003 m. Thus, the velocity of the wooden block is,

$v^2=2(9.8)(0.003)\\v=0.24\rm\; m/s$

Suppose the velocity of the bullet is v'. Thus by the law of conservation of momentum,

$m_1v=m_2v'\\10(0.24)=0.003(v')\\v'=800\rm\; m/s$

Hence, the velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.

Learn more about the conservation of momentum here;

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