A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms -1 gets embedded in it, then loss of kinetic energy will be

Question

A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms -1 gets embedded in it, then loss of kinetic energy will be (A) 4.9 J (B) 9.8 J (C) 14.7 J (D) 19.6 J

Tardigrade Admin · Accepted Answer

P i =P f ( no any external force ) 0.2 × 10=10 × v v =0.2 m / sec Loss in K.E. =(1/2) ×(0.2) × 102-(1/2) × 10(0.2)2 =(1/2) × 10 ×(0.2)[10-0.2] =9.8 J

Q. A bag of sand of mass $9.8 k g$ is suspended by a rope. A bullet of $200 g$ travelling with speed $10 m s^{- 1}$ gets embedded in it, then loss of kinetic energy will be

$4.9 J$

$9.8 J$

$14.7 J$

$19.6 J$

$P_{i} = P_{f}$ ( no any external force )
$0.2 \times 10 = 10 \times v$
$v = 0.2 m / sec$
Loss in K.E. $= \frac{1}{2} \times (0.2) \times 1 0^{2} - \frac{1}{2} \times 10 (0.2)^{2}$
$= \frac{1}{2} \times 10 \times (0.2) [10 - 0.2]$
$= 9.8 J$

Q. A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10ms−1 gets embedded in it, then loss of kinetic energy will be

4.9J

9.8J

14.7J

19.6J