Question

A bullet of mass 10 kg moving horizontally with a velocity of 400 ms strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5m

Answer 1

During the collision, apply moementum conservation (0.01)(400) + 0 = (2)V + (0.01)V'

where V = √2gh

V = √2 ×10× 0.1

V = √2 solving

V' = 120 m/sec. Ans

Answer 2

Given that,

Mass of bullet = 10 kg

Velocity = 400 m/s

Mass of block= 2 kg

Length = 5 m

Suppose, the center of gravity the block is found to rise a vertical distance of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

Let the velocity of the bullet and the block after the collision will be v₁ and v₂.

We need to calculate the velocity of bullet and block after collision

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Put the value into the formula

$10\times10^{-3}\times400+2\times0=10\times10^{-3}\times v_{1}+2\times v_{2}$.....(I)

$4=10\times10^{-3}\times v_{1}+2\times v_{2}$

We need to calculate the velocity of block

Using work energy theorem

$W=\Delta KE$

$mgh=\dfrac{1}{2}mv_{1}^2$

Put the value into the formula

$2\times10\times0.1=\dfrac{1}{2}\times2\times v_{1}^2$

$v_{1}^2=2$

$v_{1}=\sqrt{2}m/s$

We need to calculate the velocity of bullet

Putting the value of v₁ in equation (I)

$4=10\times10^{-3}\times\sqrt{2}+2\times v_{2}$

$v_{2}=\dfrac{4-10\times10^{-3}\times\sqrt{2}}{2}$

$v_{2}=1.9\ m/s$

Hence, The velocity of block and bullet after collision is √2 m/s and 1.9 m/s.