Verified answer

I am assuming that the speed is 2.30 x 10^2 m/s.

(a) Since the bullet becomes embedded, the final speed of the bullet is zero. We see that:

∆KE = -KE(initial) = (-1/2)mv^2 = (-1/2)(0.002)(2.30 x 10^2)^2 = -52.9 J.

Assuming that all of the kinetic energy gets converted into thermal energy, the ice absorbs 52.9 J of thermal energy. Since this is a small amount, we can neglect this.

Assuming that ice block is large is large, the ice will cool the bullet to 0°C

(b) Note that:

∆Q(ice) = -∆Q(bullet) ==> m(ice)*H(f) = -m(bullet)C(lead)(0 - 28.7).

Solving for m(ice) yields:

m(ice) = [-m(bullet)C(lead)(0 - 28.7)]/H(f)

= [(-0.002)(130)(0 - 28.7)]/(3.34 x 10^5)

= 2.23 x 10^-5 kg.

I hope this helps!

This question is quite peculiar. Like what is 2.30 102m/s. And is there any air resistance.

Neglecting air resistance and taking the speed to be 102m/s,

You get the Kinetic energy of the bullet, kinetic energy = (1/2)(mass)(velocity)^2

(0.5)(0.002kg)(102)^2= 10.404J

and when it stops in the ice, all the kinetic energy is converted to heat energy and the heat energy is 10.404J.

Okay, the heat formula is Q = m*c* (delta) t <--- [Q is the specific heat, M is the mass, C is the quantity of heat energy needed to raise the temperature by 1 degree and (delta) t is the change in temperature]

I cannot continue as the question did not provide the heat capacity(C) of the lead bullet.