Question

 a bullet of mass 5g on hitting a sandbag suspended in air gets embedded in it. If the velocity of the bullet be 900 m/s just before it hits the bag. Find out the speed with which the sandbag will move due to the impact of the bullet. The mass of the sandbag is 50 kg.

Answer 1

momentum of bullet before collision : 0.005 * 900 m/s = 4.5 kg-m/s
momentum of bag before collision = 0
After collision the momentum of bullet + bag is same in the same direction as the total momentum is conserved.
Momentum of bullet + bag = 50.005 * V = 4.5
         V = 0.0899 m/s

Answer 2

Answer:

Explanation:

Solution :-

Given :-

Mass of the bullet = 5 g

= 5 × 10⁻³ kg

Initial velocity of the bullet = 900 m/s

Mass of the sand bag = 50 kg

Final velocity of the bullet = 0 m/s

Initial velocity of the bag = 0 m/s

To Calculate :-

Final velocity of the bag = ??

Formula to be used :-

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Putting the values, we get

(5 ×  10⁻³ × 900) + (50 × 0) = 5 × 10⁻³ × 0) + (50 × v₂)

(45 × 10⁻¹) + 0 = 0 + 50v₂

v₂ = 45 × 10⁻¹/50

v₂ = 0.09 m/s

Hence, The mass of the sandbag is 50 kg is 0.09 m/s.