Worked examples · Sheet 1

Impulses and average forces

Theory used: Momentum Impulse of a force Impulse-momentum theorem Average force Impulsive forces

Exercise 1

A ball of mass $m = 0.45 kg$ hits a wall perpendicularly with speed $12 m/s$ and bounces back along the same direction, in the opposite sense, with speed $10 m/s$ . The contact with the wall lasts $Δ t = 20 ms$ . Compute the impulse exerted by the wall on the ball and the average force of the wall.

x axis pointing out of the wall: v_{1} = - 12 m/s, v_{2} = 10 m/s

J_{x} = Δ p_{x} = m (v_{2} - v_{1}), F_{m} = \frac{J _{x}}{Δ t}

J_{x} = 0.45 kg \cdot (10 - (- 12)) m/s = 0.45 \cdot 22 kg \cdot m/s \approx 9.9 N \cdot s

F_{m} = \frac{9.9 N \cdot s}{0.020 s} \approx 5.0 \cdot 1 0^{2} N

The problem asks for a force that acts for a short time and whose time dependence is unknown, so we use the impulse-momentum theorem, according to which the impulse of the resultant equals the change in momentum. The motion takes place along a line, therefore one component is enough; the $x$ axis is oriented from the wall outward, so the velocity before the collision is negative and the one after is positive. The change in velocity is $22 m/s$ , that is the sum of the two magnitudes, because the ball reverses the sense of its motion.

During the contact the weight also acts; it is vertical and has impulse $m g Δ t \approx 0.088 N \cdot s$ , negligible compared with $9.9 N \cdot s$ ; the weight is a non-impulsive force. Therefore the horizontal impulse is due to the wall. Dividing by the duration of the contact we obtain the average force, about $5.0 \cdot 1 0^{2} N$ , directed from the wall outward.

As a check, the average force is about $110$ times the weight of the ball, which is $4.4 N$ , so during the contact the weight is indeed negligible.

Exercise 2

A cart of mass $m = 20 kg$ moves on a horizontal track with negligible friction, with velocity $v_{1} = 1.5 m/s$ . From $t = 0$ a person pushes it in the direction of motion with a horizontal force that increases linearly from $0$ to $40 N$ in $1.0 s$ and then decreases linearly to $0$ in another $2.0 s$ . Compute the impulse of the force, the final velocity of the cart and the average force over the $3.0 s$ of the push.

Force applied to the cart (in newtons) as a function of time (in seconds): the impulse is the area of the triangle, equal to that of the dashed rectangle of height

F_{m} = 20 N

J = \int_{0}^{3.0 s} F (t) d t = \frac{1}{2} Δ t F_{m a x}, m v_{2} - m v_{1} = J ⟹ v_{2} = v_{1} + \frac{J}{m}

J = \frac{1}{2} \cdot 3.0 s \cdot 40 N = 60 N \cdot s

v_{2} = 1.5 m/s + \frac{60 N \cdot s}{20 kg} = 1.5 m/s + 3.0 m/s = 4.5 m/s

F_{m} = \frac{J}{Δ t} = \frac{60 N \cdot s}{3.0 s} = 20 N

The force has a fixed direction and variable magnitude, so the magnitude of its impulse is the area under the graph of $F (t)$ . The graph is a triangle with base $3.0 s$ and height $40 N$ , therefore the impulse is $60 N \cdot s$ ; the integral is computed without writing the expression of the force on the two segments.

The track is horizontal and frictionless, so the weight and the reaction of the track are vertical and balance each other; the only force with a horizontal component is the push. By the impulse-momentum theorem the push increases the momentum of the cart by $60 kg \cdot m/s$ , and dividing by the mass we obtain an increase in velocity of $3.0 m/s$ . The cart therefore goes from $1.5 m/s$ to $4.5 m/s$ .

The average force is the constant force that would give the same impulse in the same time. For a triangular profile it equals half the maximum value, as shown by the dashed rectangle in the figure, which has the same area as the triangle.

Exercise 3

A ball of mass $m = 0.20 kg$ is dropped from rest from a height $h_{1} = 1.25 m$ above the floor, bounces and rises back to the height $h_{2} = 0.80 m$ . The contact with the floor lasts $Δ t = 10 ms$ . Neglecting air resistance, compute the total impulse received by the ball during the contact and the average force exerted by the floor.

y axis upward: v_{1} = - 2 g h_{1}, v_{2} = 2 g h_{2}

(N_{m} - m g) Δ t = m (v_{2} - v_{1}) ⟹ N_{m} = \frac{m ( 2 g h _{2} + 2 g h _{1} )}{Δ t} + m g

2 g h_{1} = 2 \cdot 9.81 \cdot 1.25 m/s \approx 4.95 m/s, 2 g h_{2} = 2 \cdot 9.81 \cdot 0.80 m/s \approx 3.96 m/s

Δ p_{y} = 0.20 kg \cdot (3.96 + 4.95) m/s \approx 1.78 N \cdot s

N_{m} = \frac{1.78 N \cdot s}{0.010 s} + 0.20 \cdot 9.81 N \approx 178 N + 2 N \approx 1.8 \cdot 1 0^{2} N

The motion is divided into three phases. During the fall and the rise the air is negligible and the only force is the weight, which is conservative; therefore mechanical energy is conserved, and from $m g h = \frac{1}{2} m v^{2}$ we obtain the speeds immediately before and immediately after the contact. With the $y$ axis upward, the velocity before the contact is negative and the one after is positive.

During the contact the reaction of the floor, upward, and the weight, downward, act on the ball. The impulse-momentum theorem concerns the resultant, so the total impulse $1.78 N \cdot s$ is the impulse of $N - m g$ ; the average force of the floor is obtained by dividing by $Δ t$ and adding the weight. The floor exerts an average force of about $1.8 \cdot 1 0^{2} N$ , directed upward.

The weight contributes about $2 N$ out of $180 N$ , that is about one percent. This confirms that during a short collision non-impulsive forces can be neglected; when the contact lasts longer their contribution grows, and it must be included as in this calculation.