Question

2. A bullet of mass 0.2 kg, moving with a velocity of 200 ms !, strikes a stationary wooden target of mass 5 kg.
If all the energy is transferred to the wooden target, calculate the velocity with which the target moves in the
forward direction.​

Answer 1

Answer:

• The velocity from which the target will move is 40 m/s

Given:

1. Mass of bullet (M₁) = 0.2 Kg
2. Velocity of bullet (v₁) = 200 m/s
3. Mass of target (M₂) = 5 Kg
4. Velocity of target (v₂) = ?

Explanation:

$\rule{300}{1.5}$

Here we need to Apply Law of conservation of energy.

$\displaystyle \dashrightarrow \sf \underbrace{\sf \dfrac{1}{2}\;M_1\;v_1^2}_{\sf Energy\;of\;Bullet}\;=\; \underbrace{\sf \dfrac{1}{2}\;M_2\;v_2^2}_{\sf Energy\;of\;target}$

Here, we have applied the Law of conservation of energy, As the complete Energy is transferred, The Kinetic Energy of bullet equals  the Kinetic Energy of the target.

Substituting the values,

$\displaystyle \dashrightarrow\sf \dfrac{1}{2}\;M_1\;(v_1)^2\;=\;\dfrac{1}{2}\;M_2\;(v_2)^2\\\\\\\dashrightarrow\sf \dfrac{1}{2}\times 0.2 \times(200)^2=\dfrac{1}{2}\times 5 \times(v_2)^2\\\\\\\dashrightarrow\sf \cancel{\dfrac{1}{2}}\times 0.2 \times(200)^2=\cancel{\dfrac{1}{2}}\times 5 \times(v_2)^2\\\\\\\dashrightarrow\sf 0.2 \times 4\times 10^4=5 \times(v_2)^2\\\\\\\dashrightarrow\sf (v_2)^{2}=\dfrac{0.8\times 10^{4}}{5}\\\\\\\dashrightarrow\sf (v_2)^{2}=\dfrac{80\times 10^{2}}{5}$

$\displaystyle \dashrightarrow\sf (v_2)^{2}=16\times 10^{2}\\\\\\\dashrightarrow\sf v_2=\sqrt{16\times 10^{2}}\\\\\\\dashrightarrow\sf v_2=4\times 10\\\\\\\dashrightarrow\sf v_2=40\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\sf v_2=40\;m/s}}}}$

∴ The velocity from which the target will move is 40 m/s.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

The velocity from which the target will move is 40 m/s