Question

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly :

(1) 0 kg m/s 

(2) 4.2 kg m/s 

(3) 2.1 kg m/s 

(4) 1.4 kg m/s

Answer 1

Option : (2) 4.2 kg m/s

Velocity just before striking the ground

If it reaches the same height,speed remains same after collision only the direction changes.

Answer 2

Correct option is (2) 4.2 kg m/s

Given that:

Mass of ball = 0.15 kg

Height from which ball is dropped = 10 m

Impulse,  = Change in linear momentum = \(Δ\vec P\)

Velocity of ball at ground (v) = \(\sqrt{2gh}\)

\(= \sqrt{2\times 10 \times 10}\)

\(= 10 \sqrt 2 \,m/s\)

\(\vec I = 0.15 \times 10\sqrt 2(\hat j) - 0.15 \times (-10\sqrt2(\hat j))\)

\(\vec I = 0.15 \times 10\sqrt 2(\hat j) + 0.15 \times 10\sqrt 2(\hat j)\)

⇒ Magnitude of impulse = 4.2 kg m/s