IMO Practice Test — Laws of Motion
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
A 0.2 kg ball hits a wall at 10 m/s and rebounds at 10 m/s. The magnitude of the change in momentum is:
$0$
$2\ \text{kg}\cdot\text{m/s}$
$4\ \text{kg}\cdot\text{m/s}$
$20\ \text{kg}\cdot\text{m/s}$
Explanation: $\Delta p=m(v-u)=0.2(10-(-10))=4\ \text{kg}\cdot\text{m/s}$.
Question 2 of 12
Two masses 4 kg and 6 kg are connected over a frictionless pulley (Atwood machine). The acceleration is: ($g=10\ \text{m/s}^2$)
$1\ \text{m/s}^2$
$2\ \text{m/s}^2$
$4\ \text{m/s}^2$
$10\ \text{m/s}^2$
Explanation: $a=\frac{(6-4)g}{6+4}=\frac{2\times10}{10}=2\ \text{m/s}^2$.
Question 3 of 12
A block on a $37^{\circ}$ incline has $\mu_k=0.25$. Its acceleration down the slope is: ($g=10$, $\sin37^{\circ}=0.6$, $\cos37^{\circ}=0.8$)
$2\ \text{m/s}^2$
$4\ \text{m/s}^2$
$6\ \text{m/s}^2$
$8\ \text{m/s}^2$
Explanation: $a=g(\sin\theta-\mu_k\cos\theta)=10(0.6-0.25\times0.8)=10(0.4)=4\ \text{m/s}^2$.
Question 4 of 12
A 1000 kg car rounds a level curve of radius 50 m. If $\mu_s=0.5$, the maximum safe speed is: ($g=10$)
$5\ \text{m/s}$
$\sqrt{250}\ \text{m/s}$
$25\ \text{m/s}$
$50\ \text{m/s}$
Explanation: $v_{max}=\sqrt{\mu_s rg}=\sqrt{0.5\times50\times10}=\sqrt{250}\approx15.8\ \text{m/s}$.
Question 5 of 12
A bullet of 0.05 kg leaves a 5 kg gun at 200 m/s. The recoil speed of the gun is:
$1\ \text{m/s}$
$2\ \text{m/s}$
$4\ \text{m/s}$
$10\ \text{m/s}$
Explanation: $v_g=\frac{m_b v_b}{m_g}=\frac{0.05\times200}{5}=2\ \text{m/s}$.
Question 6 of 12
If the speed of a body in circular motion is doubled (radius fixed), the centripetal force becomes:
double
half
four times
unchanged
Explanation: $F=\frac{mv^2}{r}$; doubling $v$ makes $F$ four times.
Question 7 of 12
A 2 kg body moving at 3 m/s collides head-on and sticks to a 4 kg body at rest. The common velocity is:
$0.5\ \text{m/s}$
$1\ \text{m/s}$
$1.5\ \text{m/s}$
$2\ \text{m/s}$
Explanation: $v=\frac{2\times3+4\times0}{6}=1\ \text{m/s}$.
Question 8 of 12
A force-time graph gives a force of 4 N for 3 s then 0 N. The impulse delivered is:
$4\ \text{N}\cdot\text{s}$
$7\ \text{N}\cdot\text{s}$
$12\ \text{N}\cdot\text{s}$
$0\ \text{N}\cdot\text{s}$
Explanation: Impulse is the area under the graph: $4\times3=12\ \text{N}\cdot\text{s}$.
Question 9 of 12
For a conical pendulum the angle with the vertical satisfies:
$\sin\theta=\frac{v^2}{rg}$
$\tan\theta=\frac{v^2}{rg}$
$\cos\theta=\frac{v^2}{rg}$
$\tan\theta=\frac{rg}{v^2}$
Explanation: The horizontal tension component gives the centripetal force, leading to $\tan\theta=\frac{v^2}{rg}$.
Question 10 of 12
A lift of mass 200 kg is pulled up with acceleration $2\ \text{m/s}^2$. The tension in the cable is: ($g=10$)
$1600\ \text{N}$
$2000\ \text{N}$
$2400\ \text{N}$
$400\ \text{N}$
Explanation: $T=m(g+a)=200\times12=2400\ \text{N}$.
Question 11 of 12
A block just begins to slide on an incline at $30^{\circ}$. The coefficient of static friction is:
$\frac{1}{\sqrt{3}}$
$\sqrt{3}$
$\frac{1}{2}$
$1$
Explanation: $\mu_s=\tan30^{\circ}=\frac{1}{\sqrt{3}}$.
Question 12 of 12
Three blocks 1 kg, 2 kg, 3 kg in contact on a frictionless floor are pushed by 12 N on the 1 kg block. The acceleration is:
$1\ \text{m/s}^2$
$2\ \text{m/s}^2$
$4\ \text{m/s}^2$
$6\ \text{m/s}^2$
Explanation: $a=\frac{F}{m_{total}}=\frac{12}{6}=2\ \text{m/s}^2$.