Question

29. A bullet of mass 10 g moving with a speed of
20 m/s hits an ice block of mass 990 g kept on a
frictionless floor and gets stuck in it. How much ice
will melt if 50% of the lost KE goes to ice ? (initial
temperature of the ice block and bullet = 0°C)
(1) 0.001 g
(2) 0.002 g
(3) 0.003 g
(4) 0.004 g​

Answer 1

Answer:

Option (3) 0.003g

Explanation:

$\huge\underline{\underline{\ Question}}$

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A bullet of mass 10 g moving with a speed of 20 m/s hits an ice block of mass 990 g kept on a frictionless floor and gets stuck in it. How much ice

will melt if 50% of the lost K.E goes to ice ? (initial temperature of the ice block and bullet = 0°C)

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$\huge\underline{\underline{\ Answer}}$

$\huge \: given \: that = >$

$mass \: of \: bullet(m1) = 10g \\ velocity \: of \: bullet \: (u1) = 20m {s}^{ - 1} \\ mass \: of \: ice(m2) = 990g \\latent \: heat \: of \: ice = 334$

Ohk,

So suppose =>

Bullet has hit the ice already,

Now,

By law of conservation of momentum =>

It's velocity will be,

m1 × u1 = m2 × u2

10 × 20 = (990+10) × u2

u2 = 0.2 m/s

Now,

K. E(1)

= 1/2 mv^2

= 1/2 (10) (20) ^(2)

K. E(2)

= 1/2 mv^2

= 1/2 (1000) (0.2) ^(2)

K. E (loss)

= K. E (1) - K. E (2)

= 1/2 [10×400 - (1000×0.04)]

= 1.98 J

Here 50℅ of K. E goes to ice =>

So, it will be

1/2 × 1.98

0.99 J

We know that=>

Q = mL

m = Q/L

= 0.99/334

= 0.003 g

So, option (3) 0.003 g is correct ✅