Question

✪ Physics ✪
A bullet of mass 0.02 kg travelling with velocity 250 m/s strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move toghether and come to rest after travelling a distance of 40 m. The coefficient of sliding friction on the rough surface is (g = 9.8 m/s²).

✍︎ Proper steps needed ✔︎​

Answer 1

Given :

Mass of bullet = 0.02kg

Initial velocity of bullet = 250m/s

Mass of block = 0.23kg

Initial velocity of block = zero

Distance covered by combined mass before coming to rest = 40m

To Find :

The coefficient of friction between combined mass and horizontal rough surface.

Solution :

❖ First of all we need to find final velocity of the combined mass.

Since no external force acts on the system, linear momentum is conserved.

$\sf:\implies\:m_1u_1+m_2u_2=(m_1+m_2)v$

» u denotes initial velocity

» v denotes final velocity of combined mass

» m₁ denotes mass of bullet

» m₂ denotes mass of block

By substituting the given values;

$\sf:\implies\:(0.02\times250)+(0.23\times0)=(0.02+0.23)v$

$\sf:\implies\:5+0=0.25v \\ \sf:\implies\:v=\dfrac{5}{0.25} \\ \sf:\implies\:v=20\:ms^{-1}$

After the collision, combined mass starts to move together. During the motion, force due to friction opposes the motion of object and finally object comes to rest. ^^"

➙ F = mass (M) × acceleration (a)

F denotes friction force

M is the total mass

➙ μ × N = M × a

➙ μ × (Mg) = M × a

➙ a = μ g

Here we will take negative acceleration as velocity decreases with time.

Applying third equation of kinematics :p

$\sf:\implies\:v'^2-v^2=2ad \\ \sf:\implies\:0^2-20^2=2(-\mu g)(40) \\ \sf:\implies\:\mu=\dfrac{400}{2\times 10\times 40}$

$\sf:\implies\:\mu=\dfrac{400}{800}$

$\sf:\implies\:\mu=\dfrac{1}{2} \\ :\implies\:\underline{\boxed{\bf{\orange{\mu=0.5}}}}$

Answer 2

Please mark me as brainliest.