Question

A bullet with mass 20 grams and velocity 100 m/s collides with a wooden block of mass 2 kg. The wooden block is initially at rest, and is connected to a spring with k = 800 N/m. The other end of the spring is attached to an immovable wall. What is the maximum compression of the spring?

Answer 1

Answer:4.97 cm

Explanation:

Given

mass of bullet=20 gm

velocity of bullet=100 m/s

mass of block=2 kg

spring constant k=800 N/m

conserving momentum

$0.02\times 100=(0.02+2)\cdot v$

v=0.99 m/s

Now using conservation of energy

$\frac{kx^2}{2}=\frac{(2+0.02)v^2}{2}$

$\frac{800\times x^2}{2}=\frac{2.02\times 0.99^2}{2}$

x=0.0497 m

x=4.97 cm

Answer 2

We have that for the Question, it can be said that the  maximum compression of the spring will be

x=0.049m

From the question we are told

A bullet with mass 20 grams and velocity 100 m/s collides with a wooden block of mass 2 kg. The wooden block is initially at rest, and is connected to a spring with k = 800 N/m. The other end of the spring is attached to an immovable wall. What is the maximum compression of the spring?

Generally the equation for the conservation  is mathematically given as

$\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\Therefore\\\\20*10^{-3}(100)=(20*10^{-3}+2)V\\\\V=0.99m/sec$

Hence

The  maximum compression of the spring will be

$x^2=\frac{20*10^{-3}(0.99^2)}{800}\\\\x=0.049m$

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