Question

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms-1and 1 ms-1. respectively​

Answer 1

Answer:

Explanation:

Answer:

1.165 m/s

Explanation:

Mass of first objects, m₁= 100 g = 0.1 kg Mass of second object, m₂ = 200 g = 0.2 kg

Velocity of m, before collision, v₁= 2 m/s

Velocity of m₂ before collision, V₂ = 1 m/s Velocity of m₁ after collision, V3 = 1.67 m/s

after collision = V4

Velocity of m2 According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

+ m₂V2 = M₁V3+ M₂V4

→ 0.1 * 2 + 0.2*1 = 0.1 * 1.67 + 0.2 * V4

→ 0.4 0.67 + 0.2 * V4

→ V4 = 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Answer 2

Full question: Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Provided that:

• Mass of object 1st = 100 g

• Mass of object 2nd = 200 g

• Initial velocity of 1st object = 2 m/s

• Initial velocity of 2nd object = 1 m/s

• Final velocity of 1st object = 1.67 m/s

To calculate:

• Final velocity of object 2nd

Solution:

• Final velocity of object 2nd = 1.15 m/s

Knowledge required:

• SI unit of velocity = m/s
• SI unit of time = sec
• SI unit of mass = kg

Using concepts:

• Law of conservation
• Formula to convert g-kg

Using formulas:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

${\small{\underline{\boxed{\sf{\rightarrow \: 1 \: g \: = \dfrac{1}{1000} \: kg}}}}}$

Required solution:

~ Firstly let us convert g into kg!

Converting 100 grams first.

$:\implies \sf 1 \: g \: = \dfrac{1}{1000} \: kg \\ \\ :\implies \sf 100 \: g \: = \dfrac{100}{1000} \: kg \\ \\ :\implies \sf 100 \: g \: = \dfrac{1}{10} \: kg \\ \\ :\implies \sf 100 \: g \: = 0.1 \: kg \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Converting 200 grams now.

$:\implies \sf 1 \: g \: = \dfrac{1}{1000} \: kg \\ \\ :\implies \sf 200 \: g \: = \dfrac{200}{1000} \: kg \\ \\ :\implies \sf 200 \: g \: = \dfrac{2}{10} \: kg \\ \\ :\implies \sf 100 \: g \: = 0.2 \: kg \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Henceforth,

• Mass of 1st ob. = 0.1 kg
• Mass of 2nd ob. = 0.2 kg

~ Now by using law of conservation let us find out the final velocity of the second object, just we have to put the values and further solve!

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 0.1(2) + 0.2(1) = 0.1(1.67) + 0.2(v_B) \\ \\ :\implies \sf 0.2 + 0.2 = 0.167 + 0.2v_B \\ \\ :\implies \sf 0.4 = 0.167 + 0.2v_B \\ \\ :\implies \sf 0.4 - 0.167 = 0.2v_B \\ \\ :\implies \sf 0.23 = 0.2v_B \\ \\ :\implies \sf \dfrac{0.23}{0.2} \: = v_B \\ \\ :\implies \sf 1.15 \: = v_B \\ \\ :\implies \sf v_B \: = 1.15 \: ms^{-1} \\ \\ :\implies \sf Final \: velocity \: = 1.15 \: ms^{-1}$