Question

15. A 0.1 kg steel ball falls from a height of 10 m on the
ground and rebounds to a height of 7 m. (1) Why
does it not rebound to its original height? (ii) If the
dissipated energy remains in the ball in the form of
Ans. (i) Collision between the earth and the ball is
non-elastic in which energy is dissipated.
heat, then by how much will the temperature of the
ball be raised?
(ii) 0.064 °C.​

Answer 1

I Don’t Know


Increase the points

Answer 2

Explanation:

0.1 c

mass = 0.1 kg

given ball falls from a height 10 m

H1 = 10

given ball bounces back to a height of 5.4 m

H2 = 5.4

specific heat of steel = s = 460

therefore,

mg(h1 - h2) = ms tmg(h1−h2)=mst

where as t = rise in temperature

0.1 \times 10(10 - 5.4) = 0.1 \times 460 \times t0.1×10(10−5.4)=0.1×460×t

4.6 = 460 \times t4.6=460×t

t = \frac{4.6}{46}t=

46

4.6

t = 0.1 \: ct=0.1c

I hope this answer will help you

thank you