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
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
Velocity of ball after collision (vt
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
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)
Plus and minus sign indicates the opposite direction.
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.