Question

A 0.030 kg lead bullet hits a steel plate, both initially at 20?C. The bullet melts and splatters on impact. Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it.
(a)How much heat is required to increase the temperature of the lead bullet and melt it?

(b)What is the minimum speed the lead bullet must have in order to melt on impact?

(c)What happens to the remaining 20% of the bullet’s kinetic energy?

Answer 1

Answer:

(a). The required is 1871.2 J.

(b). The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Explanation:

Given that,

Mass of bullet = 0.030 kg

Temperature = 20°C

(a). We need to heat required to increase the temperature of the lead bullet and melt it

Using formula of heat

$Q=mS\Delta T+mL$

Where, m = mass of lead bullet

S = specific heat

L = latent heat

T = Temperature

Put the value into the formula

$Q=0.030\times130\times(327.5-20)+0.030\times22400$

$Q=1871.2\ J$

The required is 1871.2 J.

(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it

We need to calculate the energy

$Q=\dfrac{1871.2}{0.8}$

$Q=2339 J$

We need to calculate the speed the lead bullet

Using formula of speed

$K.E=Q$

$\dfrac{1}{2}mv^2=2339$

$v^2=\dfrac{2339\times2}{0.030}$

$v=\sqrt{\dfrac{2339\times2}{0.030}}$

$v=394.8 = 395\ m/s$

The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Hence, This is the required solution.