Question

A physics student devised the following method of measuring the muzzle velocity of a rifle: She fires a bullet into a 4.648 kg wooden block resting on a horizontal smooth surface, and attached to a spring of spring constant k = 142.7 N/m. The bullet, whose mass is 7.87 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.46 cm. What is the initial speed of the bullet?

Answer 1

Answer:

Initial velocity of the bullet will be 307.63 m /sec

Explanation:

We have given that mass of the bullet 7.87 gram m= 0.00787 kg

Mass of the block M = 4.648 kg

So total mass M+m = 4.648 +0.00787= 4.655 kg

Spring constant k = 142.7

Compression x=9.46 cm = 0.0946 m

Now according to energy conservation

$\frac{1}{2}(M+m)v^2=\frac{1}{2}kx^2$

$\frac{1}{2}\times 4.655\times v^2=\frac{1}{2}\times 142.7\times 0.0946^2$

$v=307.63m/sec$

So initial velocity of the bullet will be 307.63 m /sec

Answer 2

The initial speed of the bullet is mathematically given as

v=307.63m/sec

What is the initial speed of the bullet?

Question Parameter(s):

She fires a bullet into a 4.648 kg wooden block resting on a horizontal smooth surface

and attached to a spring of spring constant k = 142.7 N/m. The bullet, whose mass is 7.87 g.

Generally, the equation for the  energy conservation   is mathematically given as

$\frac{1}{2}(M+m)v^2=\frac{1}{2}kx^2$

Therefore

0.5*4.655*v^2=0.5* 142.7* 0.0946^2

v=307.63m/sec

In conclusion, the initial speed of the bullet

v=307.63m/sec

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