First slide

Second law

Question

A balloon has 8 g of air. A small hole is pierced into it. The air escapes at a uniform rate of 7 cm/s. If the balloon shrinks in 5 .6 second, then the average force acting on the balloon is

Easy

A

10^-4 N
B

10^-2 dyne
C

56 dyne
D

10^-6 N

Solution

The rate of change of momentum $= V \times (\frac{dm}{dt})$
$∴ Average force = \frac{7}{100} \times \frac{(8 / 1000)}{(5 \cdot 6)} = 10^{- 4} N$

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