Question

A bullet of mass 10 g and travelling at a speed of 500 m/s strikes a block of mass 2 kg which is suspended by a string of length 5 m. The bullet goes through the block in a very short time and the centre of gravity of the block is found to rise a vertical distance of 10 cm. What is the speed of the bullet just after it emerges from the block?

Answer 1

given:-

mass of bullet = 10g

initial velocity of bullet = u¹ = 500m/s

mass of block = 2000g

assumption:-

let v¹ and v² be the velocities of the blocks after collision

since the kinetic energy of the bullet is getting converted into potential energy in the absence of conservative force,

$\boxed{\red{\tt k.e = p. e }}$

$\tt \dfrac{1}{2}\cancel{m}×v^2=\cancel{m}gh$

$\tt v^2 = 10×0.1×2$

$\tt v=✓2 m/s$

_____________________________

since the bullet collides with the block,

applying law of conservation of mass,

$\boxed{\pink{\tt m^1u^1+m^2u^2=m^1v^1+m^2v^2}}$

$\tt v^1 = \dfrac{0.01×500-2×1.4}{0.01}$

$\tt v^1 = 220 m/s$

thus, the velocity of the bullet after the collision would be 220 m/s

hope this helps!!