NEET · Physics · STD 11 - 4.1 newtons laws of motion
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is ....Take \((g =9.8 m / s ^2)\)
- A 21 NS
- B 7 NS
- C 0
- D 84 NS
Answer & Solution
Correct Answer
(A) 21 NS
Step-by-step Solution
Detailed explanation
\(v_i =-\sqrt{2 gh _i}=-\sqrt{2 \times 9.8 \times 10}=-28 m / s \)
\( v _{ f } =\sqrt{2 gh _{ i }}=\sqrt{2 \times 9.8 \times 10}=14 m / s \)
\(\overrightarrow{ I }=\Delta \overrightarrow{ P }= m \overrightarrow{ v }_{ f }- m \overrightarrow{ v }_{ i }\)
\( = m \left[\overrightarrow{ v }_{ f }-\overrightarrow{ v }_{ i }\right]=0.5[14-(-28)] \)
\( =0.5 \times 42=21 NS\)
\( v _{ f } =\sqrt{2 gh _{ i }}=\sqrt{2 \times 9.8 \times 10}=14 m / s \)
\(\overrightarrow{ I }=\Delta \overrightarrow{ P }= m \overrightarrow{ v }_{ f }- m \overrightarrow{ v }_{ i }\)
\( = m \left[\overrightarrow{ v }_{ f }-\overrightarrow{ v }_{ i }\right]=0.5[14-(-28)] \)
\( =0.5 \times 42=21 NS\)
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