Question

Two identical balls ‘A’ and ‘B’ are moving with

same velocity. If velocity of‘A’ is reduced to

half and of ‘B’ to zero, then the rise in

temperatures of ‘A’ to that of ‘B’ is​

Answer 1

hi, here's your answer

For ball A:

mass= m

velocity=v

specific heat =s

Energy dissipated for ball A= Initial k.e- final k.e

= \frac{mv^{2} }{2}

2

mv

2

- m(v/2)^{2} / 2m(v/2)

2

/2

=3/8 mv²

msΔTA= 3/8 mv²

For ball b:

Energy dissipated by ball b= 1/2 m v² - 0

=1/2 m v²

msΔTb=1/2 m v²

ΔTb= v²/2s

ΔTa/ Δtb= 3v²x2s/ 8s x v²

= 3/4

∴ΔTa:ΔTb= 3:4

hope it helps

Have a great day ahead ☺☺

Answer 2

Answer:

• 3: 4

Explanation:

→For ball A:

mass= m

velocity=v

specific heat =s

Energy for ball A= Initial KE- final KE

= mv²/2

=3/8 mv²

msΔTA= 3/8 mv²

→For ball b:

Energy dissipated by ball b= 1/2 m v² - 0

=1/2 m v²

msΔTb=1/2 m v²

ΔTb= v²/2s

ΔTa/ Δtb= 3v²x2s/ 8s x v²

= 3/4

→ 3:4

$\huge\ddot{\smile}$