# Impulse – problems and solutions

1. A 0.5 kg ball free fall from a height of h1 = 7.2 meters and reflected a height of h2 = 3.2 meters. Acceleration due to gravity = 10 m/s2. Determine impulse.

Known :

Mass of ball (m) = 0.5 kg

First height (h1) = 7.2 meter

Second height (h2) = 3.2 meter

Acceleration due to gravity (g) = 10 m/s2

Wanted : Impulse (I)

Solution :

Velocity of ball before collisions (vo)

Velocity of ball before collision calculated using equation of free fall motion. Known : Height (h) = 7.2 meters, acceleration due to gravity (g) = 10 m/s2. Wanted : Final velocity after collision.

v2 = 2 g h

vo2 = 2(10)(7.2) = 144

vo = 2(10)(7.2) = 12 m/s

Velocity of ball before collision (vo) = -12 m/s. Minus sign indicates the direction of ball.

Velocity of ball after collision (vt)

Velocity of ball after collision calculated using equation of vertical motion. Known : height (h) = 3.2 meters, acceleration due to gravity (g) = -10 m/s2, final velocity at the maximum height (vt2) = 0. Wanted : Initial velocity after collision between ball and floor.

vt2 = vo2 + 2 g h

0 = vo2 + 2 (-10)(3.2)

vo2 = 64

vo = √64 = 8 m/s

Velocity of ball after collision (vt) is 8 m/s

Impulse (I)

Impulse (I) = the change in momentum (Δp)

I = m (vt – vo) = (0.5)(8-(-12)) = (0.5)(8 + 12) = (0.5)(20) = 10 Newton second

Read :  Rounding a banked curve – dynamics of circular motion problems and solutions

2. A 5-gram ball free fall from a height and strikes the floor. Acceleration due to gravity, g = 10 ms-2. Velocity of ball before collision is 6 ms-1 and after collision, the ball is reflected upright at 4 m/s. Determine the impulse.

Known :

Mass of ball (m) = 5 gram = 0.005 kg

Velocity of ball before collision (vo) = -6 m/s

Velocity of ball after collision (vt) = 4 m/s

Plus and minus sign indicates that the direction before collision is opposite with the direction after collision.

Wanted: Impulse (I)

Solution :

Impulse (I) = the total change in momentum (Δp).

I = Δp = m vt – m vo = m (vt – vo)

I = (0.005)(4 – (-6)) = (0.005)(4 + 6) = (0.005)(10) = 0.05 Newton second

Read :  Centripetal acceleration – problems and solutions

3. A 20-gram thrown with velocity of 4 m.s-1 to the left. After colliding with the wall, the ball is reflected with velocity of v2 = 2 m.s-1 to the right. Determine the impulse.

Known :

Mass of ball (m) = 20 gram = 0.020 kg Velocity of ball before collision (vo) = -4 m/s (to the left)

Velocity of ball after collision (vt) = +2 m/s (to the right)

Wanted : Impulse

Solution :

Impulse (I) = the change in momentum (Δp) = m vt – m vo

Impulse (I) = m (vt – vo) = 0.02 (2 – (-4))

Impulse (I) = 0.02 (2 + 4) = 0.02 (6)

Impulse (I) = 0.12 Newton second. 