IMO Practice Test — Motion in a Plane
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
Two equal forces give a resultant equal to one of them in magnitude. The angle between them is:
60 degrees
90 degrees
120 degrees
180 degrees
Explanation: R^2 = F^2 + F^2 + 2F^2 cos theta = F^2 gives cos theta = -1/2, so theta = 120 degrees.
Question 2 of 14
If A = i + j and B = i - j, the angle between A and B is:
0 degrees
45 degrees
90 degrees
180 degrees
Explanation: A . B = (1)(1) + (1)(-1) = 0, so the vectors are perpendicular (90 degrees).
Question 3 of 14
A projectile has the same range for angles of 15 degrees and another angle. That angle is:
30 degrees
45 degrees
75 degrees
60 degrees
Explanation: Complementary angles give equal range: 15 + 75 = 90 degrees.
Question 4 of 14
At the top of its path, a projectile launched at u and theta has speed:
0
u
u sin theta
u cos theta
Explanation: Vertical velocity is zero at the top; only the horizontal part u cos theta remains.
Question 5 of 14
The maximum height of a projectile equals its range when tan theta equals:
1
2
4
1/2
Explanation: H = R gives (u^2 sin^2 theta)/(2g) = (u^2 sin 2 theta)/g, leading to tan theta = 4.
Question 6 of 14
A body in uniform circular motion has its acceleration and velocity at an angle of:
0 degrees
45 degrees
90 degrees
180 degrees
Explanation: Velocity is tangential and centripetal acceleration points to the centre, so they are perpendicular (90 degrees).
Question 7 of 14
If the speed of a body in a circle doubles (radius unchanged), its centripetal acceleration becomes:
Double
Half
Four times
Unchanged
Explanation: a_c is proportional to v^2, so doubling v makes a_c four times larger.
Question 8 of 14
A particle moves with v = omega r. If omega is halved and r is doubled, the linear speed becomes:
Half
Double
Unchanged
Four times
Explanation: v = omega r; (omega/2)(2r) = omega r, so the speed is unchanged.
Question 9 of 14
Two projectiles are fired with the same speed at 30 and 60 degrees. The ratio of their maximum heights H(30) : H(60) is:
1 : 1
1 : 3
3 : 1
1 : 2
Explanation: H is proportional to sin^2 theta; sin^2 30 : sin^2 60 = (1/4) : (3/4) = 1 : 3.
Question 10 of 14
The magnitude of the change in velocity of a body moving with speed v through a quarter circle is:
v
2v
v sqrt(2)
0
Explanation: Initial and final velocities are perpendicular and equal in magnitude, so |delta v| = sqrt(v^2 + v^2) = v sqrt(2).
Question 11 of 14
An object thrown horizontally at 10 m/s from a 45 m cliff lands at a horizontal distance of (g = 10 m/s^2):
20 m
30 m
45 m
90 m
Explanation: t = sqrt(2 x 45/10) = sqrt(9) = 3 s; R = 10 x 3 = 30 m.
Question 12 of 14
The resultant of two vectors is maximum when the angle between them is:
0 degrees
90 degrees
180 degrees
45 degrees
Explanation: R = sqrt(A^2 + B^2 + 2AB cos theta) is maximum when cos theta = 1, i.e. theta = 0 (R = A + B).
Question 13 of 14
A car turns on a circular track of radius 50 m at 10 m/s. The centripetal acceleration is:
0.2 m/s^2
2 m/s^2
5 m/s^2
20 m/s^2
Explanation: a_c = v^2/r = 100/50 = 2 m/s^2.
Question 14 of 14
For a projectile, the velocity is minimum:
At launch
At the highest point
Just before landing
Halfway up
Explanation: At the highest point the vertical component is zero, so the speed (only u cos theta) is minimum.