IB Physics HL Theme A: Space, time and motion questions

Answer: B. Taking upward as positive, vertical motion gives $0=10+5t-5t^2$ so the positive time to reach the ground is $t=2.0\ \text{s}$ The horizontal speed is $10\cos30^\circ\ \text{m s}^{-1}$ giving a horizontal distance of $(10\cos30^\circ)(2.0)=17.3\ \text{m}$

A: uses the vertical component, $10\sin30^\circ$ as the horizontal speed.
C: uses the launch speed as the horizontal speed, omitting the cosine component.
D: doubles the flight time after finding it, adding an extra descent interval.

Remember: horizontal and vertical components determine different parts of projectile motion.

Answer: C. Translational equilibrium of the box gives $N - mg - 10 = 0$ so the platform’s force on the box is $N = (3.0)(10)+10 = 40\ \text{N}$ By Newton’s third law, the box exerts an equal-magnitude force on the platform.

A: Treats the third-law forces as cancelling, although they act on different objects, giving $0\ \text{N}$
B: Balances only the box’s weight, omitting the person’s downward force, giving $30\ \text{N}$
D: Counts the person’s downward force twice, giving $30+2(10)=50\ \text{N}$
Remember: sum forces on one object for equilibrium; third-law pairs act on different objects.

Answer: C. At $t=3.0\ \text{s}$ the graph gives $\omega=+2\ \text{rad s}^{-1}$ so the wheel rotates anticlockwise. The segment has a negative slope, so $\alpha<0$ because $\omega$ and $\alpha$ have opposite signs, the angular speed is decreasing.

A: identifies the rotation direction correctly but assumes negative angular acceleration means increasing speed.
B: reverses the stated sign convention: positive angular velocity is anticlockwise.
D: reads the angular velocity as negative; at this time it is positive, while the graph’s slope is negative.
Remember: angular speed decreases when angular velocity and angular acceleration have opposite signs.