Can you help me with this physics question? I know the final answer, but for some reason i can't get it.?

Question

A 33 g lead bullet initially at 20°C comes to rest in the block of a ballistic pendulum. Assume that half the initial kinetic energy of the bullet is converted into thermal energy within the bullet. If the speed of the bullet was 410 m/s, what is the temperature of the bullet immediately after coming to rest in the block?

The answer is 621.32 K or 350 degree C Please explain how to solve

Answer 1

KE = 1/2 mv^2
m=33m/s
v=410 m/s

KE = 1/2* 0.033kg * (410m/s)^2
=2773.65 J

1/2 of that goes to raise the temperature in the lead bullet where the specific heat is 0.1273 kJ/(kg-K)

2773.65/2 J= 1386.825 J for heating up bullet or 1.386825 kJ

Take 1.38625 kJ /0.1273(kJ/kg-K)/0.033kg = 330.1 deg K rise

or total of 20 degree C +330 deg K rise = 350 deg C

Answer 2

Seems straightforward.

Find the KE of the bullet. 1/2 of that energy went into heating up the bullet. You need to get the thermal capacity of lead.

There's a lot of units conversion involved in this problem

KE/2 = m*density*Cp*delta_T

Where Cp is the thermal capacity of lead, units joules/(m^2*K)
density kg/m^3, m kg,

And you know that KE =mv^2/2, right?

Answer 3

you want to apply the expressions pertaining to distance (correct), acceleration, time, initial and extremely last velocities. v^2 = u^2 + 2as v = u + at the position s = distance travelled u = initial speed v = very last speed t = time a = acceleration (therefore the rigidity of gravity) or shuffling issues round s = (v^2 - u^2)/2a when it comes to the ball dropped from the construction a = g = 9.8m.s^2 and u = 0, v=v when it comes to the ball thrown upward a = -g = -9.8m/s^2 (deceleration) and u = v, v = 0 ( because the rigidity of gravity is an identical yet in the option route, v will be done on a similar element from which the first ball became dropped and at the same time as at the same time as the first ball passes the point from which the 2d ball became released. you could examine this with the expression v = u + at) for the first ball distance s1 travelled s1 = (v^2 - 0^2)/2g s1 = v^2/2g Time taken is v = 0 + gt v = gt t = v/g for the 2d ball distance s2 travelled is s2 = (0^2 -v^2)/-2g s2 = -v^2/-2g or s2 = v^2/2g Time taken is 0 = v - gt v = gt t = v/g because the rigidity of gravity (g) is a continuing in this social gathering the area traveled and the time taken is only a function of the terminal speed v. subsequently at time = a million/2t both the balls would have a similar speed pondering this with the area speed courting, the area traveled by using each ball will be a similar subsequently, the balls pass one yet another at precisely the midway element of the construction.