Question

Please answer the question in the picture below.

Answer 1

Question:

• Two bodies of masses 2 kg and 3 kg moving with velocities 4 m/s and 2 m/s respectively, in the same direction collide. After collision, velocity of 2 kg body is 1 m/s in the opposite direction. Find the final velocity of 3 kg block.

Provided that:

• Mass of object one = 2 kg

• Mass of object second = 3 kg

• Initial velocity of object one = 4 mps

• Initial velocity of object second = 2 mps

• Final velocity of object one = -1 mps

Don't be confused!

• Final velocity of object one cames in negative because it goes to the opposite direction of the motion.

To calculate:

• The final velocity of 3 kg block.

Solution:

• Final velocity of 3 kg block = 16/3 mps

Using concept:

• Law of conservation of momentum

Using formula:

• Law of conservation of momentum:

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

(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.)

Required solution:

${\sf{:\implies m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}$

${\sf{:\implies 2(4) + 3(2) = 2(-1) + 3(v_B)}}$

${\sf{:\implies 8 + 6 = -2 + 3v_B}}$

${\sf{:\implies 14 = -2 + 3v_B}}$

${\sf{:\implies 14 + 2 = 3v_B}}$

${\sf{:\implies 16 = 3v_B}}$

${\sf{:\implies \dfrac{16}{3} \: = v_B}}$

• Henceforth, final velocity of object second is ${\sf{\dfrac{16}{3}}}$. Henceforth, option (b) is correct!

Answer 2

Question

Two bodies of masses 2kg and 3kg moving with velocities 4 m/s and 2m/s respectively in same direction collide. After collision, velocity of 2kg Body becomes 1 m/s in the opposite direction. Find the final velocity of 3 kg block.

Answer :-

Given that :-

• Mass of first object (m1) = 2kg
• Mass of second object (m2) = 3kg
• Initial velocity of m1 (u1) = 4m/s
• Initial velocity of m2 (u2) = 2m/s
• Final velocity of m1 = 1 m/s

To Find :-

• Final velocity of m2

Formula used :-

We will use the formula derived from "Law of conservation of momentum".

$• \: \: \sf \red{m1u1 + m2u2 = m1v1 + m2v2 }$

Where,

• m = mass of the object

• u = initial velocity of the object

• v = final velocity of the object

Now, putting the values, we get :-

$\sf \implies{m1u1 + m2u2 = m1v1 + m2v2} \: \: \\ \sf \implies{2 \times 4 + 3 \times 2 = 2 \times - 1 + 3 \times x} \\ \sf \implies{8 + 6 = - 2 + 3x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies{3x = 14 + 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies{x = \frac{16}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bf\green{ \therefore v2 = \frac{16}{3} }$

✯ Important :-

↦ What does "law of conservation of momentum" state?

Ans. "Law of conservation of momentum" states that sum of momentum before and after Collison remains unchanged, same.

↦What is "momentum"?

Ans. The product of mass and velocity is known as momentum.