IMO Practice Test — Force and Laws of Motion
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
A $0.5\ \text{kg}$ ball moving at $6\ \text{m/s}$ is stopped in $0.3\ \text{s}$. The average force is:
$1\ \text{N}$
$5\ \text{N}$
$10\ \text{N}$
$18\ \text{N}$
Explanation: $F = \Delta p/\Delta t = (0.5 \times 6)/0.3 = 3/0.3 = 10\ \text{N}$.
Question 2 of 12
Two ice skaters of masses $50\ \text{kg}$ and $75\ \text{kg}$ push off from rest. If the lighter skater moves at $3\ \text{m/s}$, the heavier one moves at:
$1.5\ \text{m/s}$
$2\ \text{m/s}$
$3\ \text{m/s}$
$4.5\ \text{m/s}$
Explanation: $50 \times 3 = 75 \times v \Rightarrow v = 150/75 = 2\ \text{m/s}$.
Question 3 of 12
A force of $20\ \text{N}$ acts for $4\ \text{s}$ on a body. The change in momentum produced is:
$5\ \text{kg m/s}$
$16\ \text{kg m/s}$
$80\ \text{kg m/s}$
$24\ \text{kg m/s}$
Explanation: $\Delta p = F \times \Delta t = 20 \times 4 = 80\ \text{kg m/s}$.
Question 4 of 12
A $1000\ \text{kg}$ car moving at $20\ \text{m/s}$ is brought to rest in $5\ \text{s}$. The braking force is:
$2000\ \text{N}$
$4000\ \text{N}$
$5000\ \text{N}$
$100\ \text{N}$
Explanation: $a = (0-20)/5 = -4\ \text{m/s}^2$; $F = ma = 1000 \times 4 = 4000\ \text{N}$.
Question 5 of 12
A $0.1\ \text{kg}$ ball hits a wall at $10\ \text{m/s}$ and bounces straight back at $10\ \text{m/s}$. The change in its momentum is:
$0\ \text{kg m/s}$
$1\ \text{kg m/s}$
$2\ \text{kg m/s}$
$10\ \text{kg m/s}$
Explanation: $\Delta p = m(v - u) = 0.1(-10 - 10) = -2\ \text{kg m/s}$, magnitude $2\ \text{kg m/s}$.
Question 6 of 12
Two equal forces act on a body in opposite directions. The body will:
accelerate
decelerate
stay in its state of motion
change direction
Explanation: Equal opposite forces are balanced (net force zero), so motion is unchanged.
Question 7 of 12
A bullet of mass $0.04\ \text{kg}$ leaves a $5\ \text{kg}$ gun at $250\ \text{m/s}$. The recoil speed of the gun is:
$1\ \text{m/s}$
$2\ \text{m/s}$
$2.5\ \text{m/s}$
$5\ \text{m/s}$
Explanation: $v_g = m_b v_b/m_g = (0.04 \times 250)/5 = 10/5 = 2\ \text{m/s}$.
Question 8 of 12
A body of mass $5\ \text{kg}$ changes its velocity from $2\ \text{m/s}$ to $7\ \text{m/s}$. Its change in momentum is:
$10\ \text{kg m/s}$
$25\ \text{kg m/s}$
$35\ \text{kg m/s}$
$45\ \text{kg m/s}$
Explanation: $\Delta p = m(v - u) = 5(7 - 2) = 25\ \text{kg m/s}$.
Question 9 of 12
If the net external force on a system is zero, then its total momentum:
increases
decreases
remains constant
becomes zero
Explanation: Conservation of momentum: total momentum stays constant.
Question 10 of 12
A $2\ \text{kg}$ object at $4\ \text{m/s}$ collides head-on and sticks to a $2\ \text{kg}$ object moving at $4\ \text{m/s}$ in the opposite direction. The common velocity is:
$0\ \text{m/s}$
$2\ \text{m/s}$
$4\ \text{m/s}$
$8\ \text{m/s}$
Explanation: Total momentum $= (2 \times 4) + (2 \times -4) = 0$, so $v = 0\ \text{m/s}$.
Question 11 of 12
Which quantity must be the same for a slow heavy truck and a fast light bullet that are equally hard to stop?
mass
velocity
momentum
acceleration
Explanation: Equal momentum makes them equally hard to stop in the same time.
Question 12 of 12
A constant force gives a $2\ \text{kg}$ body an acceleration of $6\ \text{m/s}^2$. The same force on a $3\ \text{kg}$ body gives an acceleration of:
$2\ \text{m/s}^2$
$4\ \text{m/s}^2$
$6\ \text{m/s}^2$
$9\ \text{m/s}^2$
Explanation: $F = 2 \times 6 = 12\ \text{N}$; then $a = F/m = 12/3 = 4\ \text{m/s}^2$.