Question

4. A bullet of mass 1 kg is fired and gets embedded into a block of wood of
mass I kg initially at rest. The velocity of the bullet before collision is 90m/s
a) What is the velocity of the system after collision
b) Calculate the kinetic energies before and after the collision
c) Is it an elastic collision or inelastic collision?
d) How much energy is lost in collision?

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Answer 1

Answer:

Kinetic energy = 4050 J

Final kinetic energy = 2025 J

Loss in kinetic energy = 2025 J

Explanation:

Step 1: The energy an item has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes.

Step 2:

$\begin{aligned}& \text { Mass of bullet }=1 K g \\& \text { Mass of wood }=1 K g \\& \text { velocity of bullet }=90 \mathrm{~ms}^{-1} \\& \text { By conservation o f momentum } \\& m_1 v_1+m _2 v _2=(m_ 1+m _2) V \\& v^2=0 \\& 1 \times 90=(1+1) \times V \\& V=45 m^{-1}\end{aligned}$

This the velocity of combined system

Kinetic energy$=1 / 2 \times 1 \times 90 \times 90=4050 J$

Final kinetic energy $=1 / 2 \times 45 \times 45=2025 J$

Loss in kinetic energy $=4045-2025=2025 J$

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