# A bullet of mass 10 g traveling horizontally at a speed of 100 m/s embeds itself in a block of...

## Question:

A bullet of mass 10 g traveling horizontally at a speed of 100 m/s embeds itself in a block of wood of mass 990 g suspended by strings so that it can swing freely. Find: a.the vertical height through which the block rises. B.how much of the bullet energy becomes internal energy

## Conservation of momentum and Energy:

The energy of the object will remain conserved, therefore the kinetic energy of the pendulum will convert into the potential energy of the pendulum.

Given data:

• Mass of the bullet {eq}\rm (m) = 10 \ g {/eq}
• mass of the block{eq}\rm (M) = 990 \ g {/eq}
• Initial speed of the bullet {eq}\rm (u) = 100 \ m/s {/eq}

(A)

Applyin the conservation of momentum

{eq}\rm mu = (m+M)V \\ 10 \times 100 = (10 + 990)V \\ V = 1 \ m/s {/eq}

Now, applying the energy of conservation, the kinetic energy of the block would be equal to the potential energy, therefore

{eq}\begin{align} \rm (m+M)gh &= \rm \dfrac{1}{2}(m+M)V^{2} \\ \rm 9.8 h &= \dfrac{1}{2}(1)^{2} \\ \rm h &= 0.051 \ m \\ \end{align} {/eq}

(B)

now, the intial energy of the bullet

{eq}\rm E_{1} = \dfrac{1}{2}mu^{2} \\ E_{1} = \dfrac{1}{2}(10 \times 10^{-3} \ kg) (100)^{2} \\ E_{1} = 50 \ J {/eq}

Now, the final energy of the block and bullet

{eq}\rm E_{2} = \dfrac{1}{2}(m+M)V^{2} \\ E_{2} = \dfrac{1}{2}[ (990+10)10 \times 10^{-3} \ kg] (1)^{2} \\ E_{2} = 0.5 \ J {/eq}

Now, the internal energy would be

{eq}\rm \Delta E = E_{1} - E_{2} \\ = 50 - 0.5 \\ = 49.5 \ J {/eq}