# A 5.00 g bullet moving with an initial speed of 410 \frac{m}{s} is fired into and passes through...

## Question:

A {eq}5.00 g {/eq} bullet moving with an initial speed of {eq}410 \frac{m}{s} {/eq} is fired into and passes through a {eq}1.00 kg {/eq} block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant {eq}910 \frac{N}{m} {/eq}.

(a) If the block moves {eq}5.80 cm {/eq} to the right after impact, find the speed at which the bullet emerges from the block. {eq}(\frac{m}{s}) {/eq}

(b) If the block moves {eq}5.80 cm {/eq} to the right after impact, find the energy lost in the collision. {eq}(J) {/eq}

## Conservation Of Mechanical Energy:

As we know that energy neither be created nor be destroyed but can exchange its form for example when the bullet strikes the wooden block then all the kinetic energy stored in the bullet will be imparted to the block kinetic energy and the block kinetic energy will be transferred to the attached spring and spring will absorb all the energy in the form of spring potential energy.

## Answer and Explanation:

When the bullet strikes the wooden block then all the kinetic energy of the block due to the bullet will be transferred to the spring and the spring will absorb the kinetic energy into spring potential energy:

From the conservation of energy:

{eq}K.E=P.E\\ \frac{1}{2}mv^2=\frac{1}{2}kx^2\\ \frac{1}{2}(1)(v)^2=\frac{1}{2}(910)\times (0.058)^2\\ 0.5v^2=1.530\\ v^2=3.06\\ v=1.74\ m/s {/eq}

Thus, the velocity of the block after it is struck by the bullet is 1.74 m/s.

This will be the velocity of the bullet after striking the wooden block.

Now from the conservation of momentum:

{eq}m_1u_1+m_2u_2=m_1v_1+m_2v_2\\ (0.005\times 410)+0=(0.005\times v_1)+(1\times 1.74 )\\ 2.05=(0.005\times v_1)+1.74\\ v_1=62\ m/s {/eq}

Thus, the velocity of the bullet as it emerges from the block is 62 m/s.

(B)

The initial kinetic energy of the bullet:

{eq}K.E_1=\frac{1}{2}mv^2\\ K.E_1=\frac{1}{2}(0.005)\times (410)^2\\ K.E_1=420.25\ J {/eq}

Final kinetic energy for the bullet:

{eq}K.E_2=\frac{1}{2}mv^2\\ K.E_2=\frac{1}{2}(0.005)\times (62)^2\\ K.E_2=9.61\ J {/eq}

Kinetic energy lost in spring and block system:

{eq}K.E_{sb}=\frac{1}{2}mv^2\\ K.E_{sb}=\frac{1}{2}(1)\times (1.74)^2\\ K.E_{sb}=3.0276\ J {/eq}

For energy loss:

{eq}\Delta K.E=420.25-(9.61+3.02)\\ \Delta K.E=407.62\ J {/eq}

Thus, the energy loss is 407.62 J. 