CAIE IGCSE Physics 1.6 Momentum Questions

Covered syllabus points

1.6.1—Momentum as mass × velocity 1.6.2—Impulse as force × time for which 1.6.3—Apply the principle of the 1.6.4—Resultant force as change in momentum

[Maximum number: 5]

Fig. 3.1 shows a boy throwing a ball at an object in a fairground.

Fig. 3.1

The ball has a mass of 190 g and travels horizontally with a constant speed of $6.9 \mathrm{~m} / \mathrm{s}$ .

(a)

Calculate the momentum of the ball.

momentum =

[ 2 ]

$1.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ $p=m v \text { OR }(p=) m v \text { OR } 0.19 \times 6.9 \text { OR } 190 \times 6.9$

OR $1.3 \times 10^{n}$

(b)

After hitting the object, the ball bounces back along the same straight path with a speed of $1.5 \mathrm{~m} / \mathrm{s}$ . The object has a mass of 1.8 kg .

Calculate the speed of the object after it is hit by the ball.

speed =

[ 3 ]

(speed of object =) $0.89 \mathrm{~m} / \mathrm{s}$ OR $0.88 \mathrm{~m} / \mathrm{s}$
momentum before (collision) = momentum after (collision)
$\begin{aligned} \text { OR } 1.3(\mathrm{~kg} \mathrm{~m} / \mathrm{s})=-(0.19 \times 1.5)(\mathrm{kg} \mathrm{~m} / \mathrm{s})+1.8 \mathrm{v}(\mathrm{~kg} \mathrm{~m} / \mathrm{s}) \text { OR }(\text { momentum of object }=) 1.3(\mathrm{~kg} \mathrm{~m} / \mathrm{s})+(0.19 \times 1.5)(\mathrm{kg} \mathrm{~m} / \mathrm{s}) \text { OR }(\text { momentum of object }=) 1.3(\mathrm{~kg} \mathrm{~m} / \mathrm{s})+0.29(\mathrm{~kg} \mathrm{~m} / \mathrm{s}) \text { OR }(\text { momentum of object }=) 1.6(\mathrm{~kg} \mathrm{~m} / \mathrm{s}) \end{aligned}$
(speed of object = ) $\{1.3+(0.19 \times 1.5)\} / 1.8(\mathrm{~m} / \mathrm{s})$

OR (speed of object =) 1.6 / 1.8

[Maximum number: 4]

Fig. 3.1 shows two children balanced on a seesaw. A seesaw is a length of wood which rotates about a central pivot.

Fig. 3.1

(a)

The concrete floor under the seesaw is replaced with a rubber floor. A child falls from the seesaw and experiences an impulse when they hit the floor.

[ 4 ]

(i)

Define impulse.

[ 1 ]

(impulse is the) force × time (for which the force acts) $\mathbf{OR} I=F \times t$

OR (impulse = ) change in momentum OR $\Delta\{m v\}$

(ii)

Explain how the rubber floor reduces injury to the child.

Use ideas about impulse, force, momentum and time in your answer.
[Total: 8]

[ 3 ]

any three from:

- change in momentum/impulse is the same (on both floors)

- (change in momentum/lmpulse) is over longer time

- force = rate of change of momentum OR $F=\Delta\{m v\} / \Delta t$

- less force on child (so less injury)

[Maximum number: 3]

Fig. 2.1 shows two identical trolleys, P and Q, held at rest on a frictionless horizontal surface. A load is fixed to trolley P .

Fig. 2.1

There is a compressed spring between trolley P and trolley Q .
The trolleys are released. As the spring expands, it pushes the trolleys apart.
Trolley Q moves to the right at a constant speed of $0.36 \mathrm{~m} / \mathrm{s}$ .
The mass of each trolley is 1.2 kg . The mass of the load on trolley P is 1.5 kg .
The spring has negligible mass.

Calculate:

the speed at which trolley P moves to the left
speed of P=

$0.16 \mathrm{~m} / \mathrm{s}$
conservation of momentum OR $m_{\mathrm{P}} v_{\mathrm{P}}=m_{\mathrm{Q}} v_{\mathrm{Q}}$ OR $2.7 \times v_{\mathrm{P}}=1.2 \times 0.36$ OR $\left(m_{\mathrm{Q}} v_{\mathrm{Q}}=\right) 0.432$ seen
$\left(v_{\mathrm{P}}=\right) m_{\mathrm{Q}} v_{\mathrm{Q}} / m_{\mathrm{P}}$ OR $1.2 \times 0.36 / 2.7$

All question bank results loaded

CAIE IGCSE Physics 1.6 Momentum Topic Practice

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (i)

Question (ii)

Question 3