Momentum and Impulse I
Newton's laws of motion
  1. First law: Newton's first law of motion states that every body continues in state of rest or uniform motion in a straight line unless acted upon by an external force. Inertia is the tendency for an object to remain at rest or maintain uniform speed in a straight line. The first law is also called the law of Inertia.
  2. Second law:The second law of motion states that the rate of change of momentum is directly proportional to the force applied and takes place in the direction of the force. \( F = ma \) (Law of Acceleration)
  3. Third law: Newton's third law of motion states that for every action, there is an equal and opposite reaction. (Law of Action and Reaction)

From Newton's second law; $$ F = \frac{mv-mu}{t} $$ $$ F = m(\frac{v-u}{t}) = ma $$ where change in momentum = \( mv-mu \) $$ \text{initial momentum} = mu $$ $$ \text{final momentum} = mv $$ $$ Ft = m(v-u) $$ $$ \text{But, Impulse} = Ft $$ $$ \therefore Impulse = m(v-u) $$ Hence, Impulse = change in momentum

For a body that rebounds when hit unto a stationary body, the change in momentum is given by: $$ Ft = m(v_b-v_a) $$ $$ \text{where,} $$ $$ v_b = \text{velocity before rebound} $$ $$ v_a = \text{velocity after rebound} $$

Calculations

Example 1: An external force of magnitude 100N acts on a particle of mass 0.15kg for 0.03s. Calculate the change in speed of the particle. (WAEC)

Solution

$$ f = 100N $$ $$ m = 0.15kg $$ $$ t = 0.03s $$ $$ ft = m{(v-u)} $$ $$ v-u = \frac{ft}{m} $$ $$ v-u = \frac{100 × 0.03}{0.15} $$ $$ \text{change in velocity} = 20m/s $$

Example 2: When taking a penalty kick, a footballer applies a force of 30.0N for a period of 0.05s. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off.

Solution

$$ f = 30N $$ $$ t = 0.05s $$ $$ m = 0.075kg $$ $$ u = 0 $$ $$ v - u = \frac{ft}{t} $$ $$ v - 0 = \frac{30×0.05}{0.075} $$ $$ speed(v) = 20m/s $$

Example 3: A bullet of mass 120g is fired horizontal into a fixed wooden block with a speed of 20m/s. If the bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:

  1. Magnitude of the resistance
  2. Distance moved by the bullet in the wood
    3. (WAEC)

Solution

$$ mass = 120g = 0.12kg $$ $$ u = 20m/s $$ $$ v = 0 $$ $$ t = 0.1s $$
  1. The magnitude of the force is given by: $$ f = m \frac{v-u}{t} $$ $$ f = 0.12 × \frac{(-20)}{0.1} $$ $$ a = -200m/s² $$ The negative sign shows the deceleration $$ f = 0.12 × 200 $$ $$ f = 24N $$
  2. The distance covered is given by: $$ a = -200m/s² $$ $$ v² = u² + 2as $$ $$ 0² = (-20)² + 2 × (-200) × s $$ $$ -400 = -400s $$ $$ s = \frac{-400}{-400} $$ $$ s = 1m $$

Example 4: A body of mass 5kg initially at rest is acted upon by two mutually perpendicular forces 12N and 5N. Determine the acceleration of the body.

Solution

Since the two forces are acting perpendicular to each other, we can resolve the resultant force using Pythagoras theorem

$$ F² = 12² + 5² $$ $$ F² = 144 + 25 $$ $$ F² = 169 $$ $$ F = \sqrt{169} $$ $$ F = 13N $$ $$ a = \frac{f}{m} $$ $$ a = \frac{13}{5} $$ $$ a = 2.60m/s² $$

Example 5: A ball of mass 5kg hits a smooth vertical wall normally with a speed of 2m/s and rebounds with the same speed. Determine the impulse experienced by the ball.(WAEC)

Solution

$$ m = 5kg $$ $$ \text{speed before rebound}{(v_b)} = 2m/s $$ $$ \text{speed after rebound}{(v_a)} = -2m/s $$ The speed is negative because the ball rebounds at a velocity opposite that of the velocity before rebound $$ Ft = m(v_b-v_a) $$ $$ I = 5(2-(-2) $$ $$ I = 5(2+2) $$ $$ I = 20Ns $$


Example 6: A body of mass 100g moving with a velocity of 10m/s collides with a wall. If after the collision, it moves with a velocity of 2m/s in the opposite direction, calculate the change in momentum.

Solution

$$ mass = 100g = 0.1kg $$ $$ v_b = 10m/s $$ $$ v_a = 2m/s $$ $$\text{Change in momentum} = m{(v_b-v_a)} $$ $$ = 0.1(10-(-2)) $$ $$ = 0.1 × 12 $$ $$ \text{Change in momentum} = 1.2Ns $$



Change in momentum Calculator







Formulas:
  1. For 1: \( f = ma \)

  2. For 2: \( a = \frac{F}{m} \)

  3. For 3: \( I = m(v-u) \)

  4. For 4: \( p = mv \)

  5. For 5: \( I = Ft \)

  6. For 6: \( I = m(v_b + v_a) \)

  7. For 7: \( \Delta v = \frac{Ft}{m} \)

  8. For 8: \( F = \frac{m(v-u)}{t} \)

Select the formula to use the calculator. Input the values for the parameters to solve using the calculator. Use the information below as a guide.

Summary