Same Blow, Different Contact Time
A ball of mass $0.16\ \text{kg}$ arrives at $20\ \text{m s}^{-1}$ and leaves the bat at $20\ \text{m s}^{-1}$ the other way, so its momentum changes by $0.16 \times 40 = 6.4\ \text{kg m s}^{-1}$ — that change is the shaded area and it never alters. Slide the contact time: as the contact lengthens the average force must fall, which is exactly why a fielder gives with the catch.
average force $F = \dfrac{\Delta p}{t} =$ 320 N from $6.4\ \text{N s}$ delivered over 0.020 s
The rectangle always has the same area of $6.4\ \text{N s}$: a shorter contact means a taller rectangle, a longer contact a flatter one.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. The velocity of a body of mass $2\ \text{kg}$ changes by $3\ \text{m s}^{-1}$. The change in its momentum is:
R2. A constant force of $30\ \text{N}$ acts on a body for $4\ \text{s}$. The product $Ft$ is:
R3. A particle moves at $5\ \text{m s}^{-1}$ and then at $3\ \text{m s}^{-1}$ in the opposite direction. Its change in velocity is:
A force that acts for a time delivers an impulse, and the unit is the newton second:
| Quantity | Statement |
|---|---|
| Constant force | $I = Ft$ |
| Varying force | $I$ is the area under the force–time graph |
| Unit | $\text{N s}$, identical to $\text{kg m s}^{-1}$ |
Mini-example: a force of $120\ \text{N}$ held for $0.05\ \text{s}$ delivers $120 \times 0.05 = 6\ \text{N s}$; the same area could also be produced by a gently growing force held for longer.
Newton's second law in impulse form:
| Form | Statement |
|---|---|
| Along a line | $I = mv - mu$ |
| Magnitude | $I = m(v + u)$ when the body rebounds |
Mini-example: a ball of mass $0.2\ \text{kg}$ strikes a wall at $5\ \text{m s}^{-1}$ and rebounds at $3\ \text{m s}^{-1}$. With the approach as positive, $I = 0.2\,(-3) - 0.2\,(5) = -1.6\ \text{N s}$, so the wall delivers $1.6\ \text{N s}$ away from itself.
Catching a ball, landing from a jump or hitting a crash barrier all set the change in momentum — the approach speed and the final rest. What the surface can choose is how long the contact lasts.
Since $F = \dfrac{\Delta p}{t}$, stretching the contact time lowers the average force in proportion. Mini-example: a change of $6\ \text{N s}$ delivered in $0.02\ \text{s}$ needs $300\ \text{N}$, but the same change spread over $0.5\ \text{s}$ needs only $12\ \text{N}$.
Split the contact into thin strips: each strip of height $F$ and width $\delta t$ contributes $F\,\delta t$ to the momentum change, so the whole impulse is the area under the graph.
Groups of 4: clear one stage at a time. Stage 1 finds the change in momentum when a ball rebounds off a wall, stage 2 turns that impulse into the average contact force, and stage 3 predicts the speed of a ball struck by a known force. A wrong option locks and reveals a hint, and a stage stays locked until the one above it is solved.
Work alone through the four cards. Decide for each claim whether it is correct or hides an error, then read the explanation. Two of the four claims are sound.
A ball of mass $0.5\ \text{kg}$ is dropped from a height of $0.45\ \text{m}$ and bounces to a height of $0.2\ \text{m}$. Find the change in momentum of the ball during the bounce. Take $g = 10\ \text{m s}^{-2}$.
Two heights give two speeds, using $v^{2} = 2gh$: the speed on arrival at the floor, and the speed with which the ball leaves it in the opposite direction. Take the downward direction as positive and remember that the rebound velocity is negative.
Dropped from rest: $u^{2} = 2gh = 2 \times 10 \times 0.45 = 9$, so the ball meets the floor at $u = 3\ \text{m s}^{-1}$ downwards.
Rising to $0.2\ \text{m}$: $v^{2} = 2 \times 10 \times 0.2 = 4$, so $v = 2\ \text{m s}^{-1}$ upwards.
With downwards positive, $v = -2$. Then $\Delta p = m(v - u) = 0.5\,(-2 - 3) = -2.5\ \text{kg m s}^{-1}$, so the momentum changes by $2.5\ \text{N s}$ directed upwards. The rebound adds the two speeds: $0.5(3 + 2) = 2.5$.
A ball of mass $0.25\ \text{kg}$ strikes the ground at $8\ \text{m s}^{-1}$ and rebounds at $6\ \text{m s}^{-1}$. The ball is in contact with the ground for $0.05\ \text{s}$. Find (i) the change in the ball's momentum during the bounce (ii) the average force exerted by the ground on the ball.
The bounce reverses the velocity, so the magnitude of the momentum change is the sum of the two speeds multiplied by the mass. Then divide the impulse by the contact time to get the average force.
Taking the incoming direction as positive, $\Delta p = 0.25\,(-6) - 0.25\,(8) = -3.5\ \text{kg m s}^{-1}$, so the momentum changes by $3.5\ \text{N s}$ away from the ground.
The impulse equals the change in momentum, so $F = \dfrac{3.5}{0.05} = 70\ \text{N}$, directed away from the ground.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A constant force of $24\ \text{N}$ acts on a body for $3\ \text{s}$. The impulse delivered is:
Q2. A ball receives an impulse of $6\ \text{N s}$ while in contact with a bat for $0.03\ \text{s}$. The average force on the ball is:
Q3. A ball of mass $0.4\ \text{kg}$ travels at $6\ \text{m s}^{-1}$ and rebounds at $3\ \text{m s}^{-1}$. The magnitude of the change in its momentum is:
Q4. The area under a force–time graph represents:
Work each on paper, then reveal the solution to check your working.
Ball $X$ has mass $30\ \text{g}$ and moves at $0.51\ \text{m s}^{-1}$; the direction in which $X$ is travelling is taken as positive. Ball $Y$ has mass $60\ \text{g}$ and is stationary. $X$ collides with $Y$ and, after the impact, $X$ moves at $0.01\ \text{m s}^{-1}$ in the positive direction. Ball $Y$ then hits a wall and rebounds with half the speed with which it hit the wall. Work out how much momentum was absorbed by the wall.
Reveal Solution
First the collision: $0.03 \times 0.51 = 0.03 \times 0.01 + 0.06\,v$, so $0.06v = 0.015$ and $v = 0.25\ \text{m s}^{-1}$ towards the wall. $Y$ leaves the wall at $0.125\ \text{m s}^{-1}$ in the opposite direction, so the momentum absorbed is $0.06\,(0.25 + 0.125) = 0.0225\ \text{kg m s}^{-1}$ (that is $2.25 \times 10^{-2}\ \text{N s}$).
Ball $X$ of mass $30\ \text{g}$ is moving at $0.01\ \text{m s}^{-1}$ in the positive direction when it is struck by ball $Y$ of mass $60\ \text{g}$, which is moving at $0.125\ \text{m s}^{-1}$ in the negative direction. After the impact the speed of $X$ is $0.15\ \text{m s}^{-1}$. Explain why $X$ must be travelling in the negative direction after being hit by $Y$.
Reveal Solution
The balls approach each other at $0.01 + 0.125 = 0.135\ \text{m s}^{-1}$. If $X$ ended up travelling in the positive direction, then, keeping the total momentum equal to $0.03(0.01) + 0.06(-0.125) = -0.0072\ \text{kg m s}^{-1}$, $Y$ would have to travel at $0.195\ \text{m s}^{-1}$ in the negative direction. The balls would then be separating at $0.15 + 0.195 = 0.345\ \text{m s}^{-1}$, faster than they approached — impossible, because a collision cannot manufacture energy. So the impulse from the heavier $Y$ reverses $X$: it travels in the negative direction.
Ball $X$ of mass $30\ \text{g}$ is moving at $0.01\ \text{m s}^{-1}$ in the positive direction when it is struck by ball $Y$ of mass $60\ \text{g}$, which is moving at $0.125\ \text{m s}^{-1}$ in the negative direction. After the impact $X$ moves at $0.15\ \text{m s}^{-1}$ in the negative direction. Find the final velocity of ball $Y$.
Reveal Solution
The total momentum before the impact is $0.03(0.01) + 0.06(-0.125) = -0.0072\ \text{kg m s}^{-1}$. After the impact: $0.03(-0.15) + 0.06v = -0.0072$, so $0.06v = -0.0027$ and $v = -0.045\ \text{m s}^{-1}$: ball $Y$ continues at $0.045\ \text{m s}^{-1}$ in the negative direction.
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