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Is Matter Around Us Pure

Question

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Two objects of masses 100 g and 200 g are moving in along the same line and direction with velocities of 2 ms^–1 and 1 ms^–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^–1. Determine the velocity of the second object?

Solution

$Given comma space Mass space of space one space object comma space straight m subscript 1 space equals space 100 space straight g space equals space 0.1 space kg Mass space of space second space object comma space space straight m subscript 2 space equals space 200 space straight g space equals space 0.2 space kg Initial space velocity space of space 1 to the power of st space object comma space straight u subscript 1 space equals space 2 space ms to the power of negative 1 end exponent Initial space velocity space of space 2 to the power of nd space object space straight u subscript 2 space equals space 1 space ms to the power of negative 1 end exponent Acquired space velocity space of space 1 to the power of st space object comma space straight v subscript 1 space equals space 1.67 space ms to the power of negative 1 end exponent Velocity space acquired space by space the space second space object comma space straight v subscript 2 space equals space ?$
$Now comma space According space to space the space law space of space conservation space of space momentum comma space space space space space space space space space straight m subscript 1 straight u subscript 1 space plus space straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus space straight m subscript 2 straight v subscript 2 rightwards double arrow space space space 0.1 cross times 2 space plus space 0.2 space cross times space 1 space equals space 0.1 space cross times space 1.67 space plus space 0.2 space straight v subscript 2 rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0.4 space equals space 0.167 plus 0.2 space straight v subscript 2 rightwards double arrow space space space space space space space space space space space space space space space straight v subscript 2 space equals space fraction numerator 0.4 minus 0.167 over denominator 0.2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals 1.165 space ms to the power of negative 1 end exponent.$

v₂ is the velocity of the second object.

Some More Questions From Is Matter Around Us Pure Chapter

Which separation techniques will you apply for the separation of the following?

Tea leaves from tea

Which separation techniques will you apply for the separation of the following?

Iron pins from sand.

Which separation techniques will you apply for the separation of the following?

Wheat grains from husk

Which separation techniques will you apply for the separation of the following?

Fine mud particles suspended in water.

Write the steps you would use for making tea. Use the words—solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

Substance Dissolved

Temperature in K

283

293

313

333

353

Potassium nitrate

21

32

62

106

107

Sodium chloride

36

36

36

37

37

Potassium chloride

35

35

40

46

54

Ammonium chloride

24

37

41

55

66

What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

Substance Dissolved

Temperature in K

283

293

313

333

353

Potassium nitrate

21

32

62

106

107

Sodium chloride

36

36

36

37

37

Potassium chloride

35

35

40

46

54

Ammonium chloride

24

37

41

55

66

Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

Substance Dissolved

Temperature in K

283

293

313

333

353

Potassium nitrate

21

32

62

106

107

Sodium chloride

36

36

36

37

37

Potassium chloride

35

35

40

46

54

Ammonium chloride

24

37

41

55

66

Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

Substance Dissolved

Temperature in K

283

293

313

333

353

Potassium nitrate

21

32

62

106

107

Sodium chloride

36

36

36

37

37

Potassium chloride

35

35

40

46

54

Ammonium chloride

24

37

41

55

66

What is the effect of change of temperature on the solubility of a salt?

Explain the following giving examples:

Saturated solution

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