Force and Laws of Motion — Question 17

Given,

Mass of the dumb-bell (m) = 10 kg

Distance covered (s) = 80 cm

Convert cm to m

100 cm = 1 m

So, 80 cm = $\dfrac{80}{100}$ = 0.8 m

Initial velocity (u) = 0

Acceleration (a) = 10 ms^-2

Terminal velocity (v) = ?

Momentum of the dumb-bell when it hits the ground = mv

As per the third law of motion,

v² - u² = 2as

Substituting we get,

v² - 0 = 2 x 10 x 0.8

v² = 16

v = $\sqrt{16}$ ms^-1

v = 4 ms^-1

Hence, dumb-bell strikes the ground with a velocity of 4 ms^-1

The momentum transferred by the dumb-bell to the floor = mv = 10 × 4 = 40 kgms^-1

Hence, momentum transferred = 40 kgms^-1