Momentum Worked ExamplesIGCSE Physics 0625

(a) Momentum is mass × velocity.

(b) Substitute, and give the unit kg m/s.

(c) Velocity has a direction, so momentum has a direction too: momentum is a vector. The ball's momentum is $48\text{ kg m/s}$ along the lane.

Resultant force is the change in momentum per unit time. At constant velocity (and constant mass) the momentum does not change.

A is the misconception that motion needs a resultant force. B ignores the resistive forces, which here exactly balance the forward force. D confuses a constant momentum with zero momentum: $p = mv$ is not zero because $v$ is not zero.

(a) Zero — because momentum = mass × velocity, and B's velocity is zero.

(b) Momentum is conserved: total momentum before = total momentum after. Before, only A has momentum.

After, the joined gliders move together, so use their combined mass.

Solve for the combined mass, then subtract the mass of A.

(a) The change in momentum is from $mv$ to zero.

Resultant force = change in momentum ÷ time.

(b) The change in momentum needed to stop is still $360\text{ kg m/s}$. Since $F\Delta t = \Delta(mv)$, a smaller force must act for a longer time. The time to stop increases.

Momentum is a vector. Take right as positive, so Y's velocity is $-1.5\text{ m/s}$.

After the collision the cars move together, so their masses combine.

Momentum is conserved, so set before = after and solve.

The answer is positive, so the cars move to the right — the original direction of X.

(a) Take the direction towards the wall as positive. The final velocity is $-20\text{ m/s}$ because the ball reverses.

Evaluate. The magnitude is what is usually asked for; the sign shows it is away from the wall.

(b) Convert the time to seconds: $4.0\text{ ms} = 0.0040\text{ s}$. Then force = change in momentum ÷ time.

Impulse = change in momentum. Take away from the player as positive, so the initial momentum is $-4.0\text{ kg m/s}$.

Positive means away from the player. The first $4.0\text{ N s}$ of the impulse stops the ball; the remaining $5.5\text{ N s}$ sends it back.

B is only the change in momentum. A adds the two values, ignoring the reversal of direction. C has the right size but the wrong direction.

(a) Resultant force is the change in momentum per unit time.

$F$ = resultant force, $\Delta p$ = change in momentum, $\Delta t$ = time taken for that change.

(b) In each second, $3.0\text{ kg}$ of water loses all of its horizontal velocity.

The wall exerts this force on the water; by the same amount, the water pushes on the wall.

Before separating, the skaters move as one object of mass $125\text{ kg}$.

After, the total momentum is shared between them. Momentum is conserved.

Solve for $v$.

$v$ is positive, so P still moves forwards, but more slowly than before.

(a) Speed = card length ÷ time. Convert $10.0\text{ cm}$ to $0.100\text{ m}$.

Before, only A moves. After, both trolleys move together with a combined mass of $1.60\text{ kg}$.

(b) Compare the numbers: the momentum after is $0.025\text{ kg m/s}$ less, a difference of about $4\%$.

The difference is small — within what you would expect from friction and timing uncertainty — so the results support the conclusion.

(c) Raising one end of the runway slightly compensates for friction: it is tilted just enough that a trolley, once pushed, rolls down at a constant speed, so no momentum is lost to friction between the gates.

(a) Resultant force = change in momentum ÷ time, which is the gradient of this graph. Read the axis carefully: $24$ means $24\,000\text{ kg m/s}$.

(b) The line is horizontal, so the momentum does not change. The resultant force is zero — the car moves at constant velocity (the driving force balances the resistive forces).

(c) The gradient is negative: the momentum falls from $24\,000$ to $6000\text{ kg m/s}$ in $4.0\text{ s}$.

The minus sign means the force acts against the motion (braking). For the speed, use $v = p/m$.

(a) Take right as positive. The ball's final velocity is $-1.6\text{ m/s}$. Momentum is conserved.

Solve.

(b) For the ball, find its change in momentum and divide by the contact time.

For the block, its momentum goes from $0$ to $1.2 \times 2.2 = 2.64\text{ kg m/s}$.

The forces are equal in size and opposite in direction — as expected, since the momentum lost by one object is gained by the other.

(c) Friction must remove all of the block's momentum: $\Delta t = \Delta p / F$.