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 :

First we determine the speed of the ball just before it strikes the ground. The ball is released from rest at a height $h\; =\; 10\backslash \; \backslash text\{m\}$ and falls under gravity $g\; =\; 10\backslash \; \backslash text\{m/s\}^2$. Using the energy or kinematic relation

$v\; =\; \backslash sqrt\{2\; g\; h\}$

we obtain

$v\; =\; \backslash sqrt\{2\; \backslash times\; 10\backslash \; \backslash text\{m/s\}^2\; \backslash times\; 10\backslash \; \backslash text\{m\}\}\; =\; \backslash sqrt\{200\}\backslash \; \backslash text\{m/s\}\; \backslash approx\; 14.14\backslash \; \backslash text\{m/s\}$

This speed is directed downward when the ball contacts the ground.

After the perfectly elastic rebound the ball rises to the same height, so its speed immediately after leaving the ground has the same magnitude but opposite direction (upward). Thus the final speed is also $v\; =\; 14.14\backslash \; \backslash text\{m/s\}$ upward.

The impulse $\backslash mathbf\{J\}$ delivered by the ground equals the change in momentum of the ball:

$\backslash mathbf\{J\}\; =\; \backslash Delta\; \backslash mathbf\{p\}\; =\; m\; \backslash ,\; \backslash mathbf\{v\_f\}\; -\; m\; \backslash ,\; \backslash mathbf\{v\_i\}$

Because the velocities are opposite, the magnitudes add:

$|\backslash mathbf\{J\}|\; =\; m\backslash ,(v\_f\; +\; v\_i)\; =\; m\backslash ,(v\; +\; v)\; =\; 2\; m\; v$

Substituting the given mass $m\; =\; 0.15\backslash \; \backslash text\{kg\}$ and the speed $v\; \backslash approx\; 14.14\backslash \; \backslash text\{m/s\}$:

$|\backslash mathbf\{J\}|\; =\; 2\; \backslash times\; 0.15\backslash \; \backslash text\{kg\}\; \backslash times\; 14.14\backslash \; \backslash text\{m/s\}\; \backslash approx\; 4.242\backslash \; \backslash text\{kg·m/s\}$

Rounded to two significant figures, the magnitude of the impulse is

$|\backslash mathbf\{J\}|\; \backslash approx\; 4.2\backslash \; \backslash text\{kg·m/s\}$

Hence the correct answer is **4.2 kg m/s**.