Question

Two bullets have masses of 3.0 g and 6.0 g respectively. Both are fired with a speed of 40.0 m/s.(a) Which bullet has more kinetic energy? (b) what is the ratio of their kinetic energies?

Answer 1

Answer:

(a) The bullet with a mass of 6.0 g

(b) $\frac{K_2}{K_1}=2$

Explanation:

(a) Kinetic energy is defined as:

$K=\frac{mv^2}{2}$

So, we have:

$K_1=\frac{3*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_1=2.4J$

$K_2=\frac{6*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_2=4.8J$

The bullet with a mass of 6.0 g have more kinetic energy

(b) We have $m_2=2m_1(1)$ and $v_1=v_2(2)$. So:

$K_1=\frac{m_1v_1^2}{2}\\K_2=\frac{m_2v_2^2}{2}\\\frac{K_2}{K_1}=\frac{\frac{m_2v_2^2}{2}}{\frac{m_1v_1^2}{2}}(3)$

Replacing (1) and (2) in (3):

$\frac{K_2}{K_1}=\frac{\frac{2m_1v_1^2}{2}}{\frac{m_1v_1^2}{2}}\\\frac{K_2}{K_1}=2$

Answer 2

$m_2(6 \ g)$ bullet has more K.E and the ration will be "$K_2 = 2K_1$".

Given:

Mass,

• $m_1 = 3 \ g$
• $m_2 = 6 \ g$

Speed,

• $v = 40 \ m/s$

The Kinetic energy will be:

→ $K_1 = \frac{1}{2} m_1 v^2$

        $= \frac{1}{2}\times 3\times (40)^2$

        $= 2400 \ J$

and,

→ $K_2 = \frac{1}{2} m_2 v^2$

        $= \frac{1}{2}\times 6\times (40)^2$

        $= 4800 \ J$

We can see that "$m_2$" is more than "$m_1$".

hence,

The ratio will be:

→ $\frac{K_2}{K_1} = \frac{4800}{2400}$

        $= 2$

  $K_2 = 2 K_1$

Thus the above approach is right.

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