Online Test — Motion in a Plane
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Which of the following is a scalar quantity?
Velocity
Displacement
Mass
Force
Explanation: Mass has only magnitude, so it is a scalar.
Question 2 of 20
Two forces of 3 N and 4 N act at right angles. The resultant is:
1 N
5 N
7 N
12 N
Explanation: R = sqrt(3^2 + 4^2) = sqrt(25) = 5 N.
Question 3 of 20
The y-component of a vector A making angle theta with the x-axis is:
A cos theta
A sin theta
A tan theta
A
Explanation: The component along the y-axis is A sin theta.
Question 4 of 20
The magnitude of a unit vector is:
0
1
Equal to the vector
Depends on direction
Explanation: A unit vector always has magnitude 1.
Question 5 of 20
The path of a projectile (ignoring air resistance) is a:
Circle
Straight line
Parabola
Ellipse
Explanation: Eliminating time gives y as a quadratic in x, so the path is a parabola.
Question 6 of 20
The horizontal range of a projectile is maximum at a projection angle of:
30 degrees
45 degrees
60 degrees
90 degrees
Explanation: R = u^2 sin(2 theta)/g is maximum when sin(2 theta) = 1, i.e. theta = 45 degrees.
Question 7 of 20
During projectile motion, the horizontal velocity component:
Increases
Decreases
Stays constant
Is zero at the top
Explanation: There is no horizontal acceleration, so u cos theta is constant throughout.
Question 8 of 20
At the highest point of a projectile's path, its vertical velocity is:
Maximum
Zero
Equal to u
Negative
Explanation: At maximum height the vertical motion momentarily stops, so the vertical velocity is zero.
Question 9 of 20
A ball is projected at 20 m/s at 30 degrees. Its time of flight (g = 10 m/s^2) is:
1 s
2 s
4 s
0.5 s
Explanation: T = 2u sin theta/g = 2 x 20 x 0.5 / 10 = 2 s.
Question 10 of 20
A projectile fired at 40 m/s at 45 degrees has a range (g = 10 m/s^2) of:
80 m
120 m
160 m
320 m
Explanation: R = u^2 sin(90)/g = 1600 x 1 / 10 = 160 m.
Question 11 of 20
In uniform circular motion, the quantity that remains constant is:
Velocity
Speed
Acceleration direction
Direction of motion
Explanation: Only the speed is constant; the velocity changes because the direction changes.
Question 12 of 20
The centripetal acceleration of a body moving at speed v in a circle of radius r is:
v r
v^2 / r
r / v
v / r^2
Explanation: a_c = v^2 / r = omega^2 r, directed towards the centre.
Question 13 of 20
The relation between linear speed and angular velocity is:
v = omega / r
v = omega r
v = r / omega
v = omega^2 r
Explanation: Linear speed v = omega r.
Question 14 of 20
The centripetal force on a body moving in a circle is directed:
Along the tangent
Towards the centre
Away from the centre
Opposite to velocity
Explanation: Centripetal force always points towards the centre of the circle.
Question 15 of 20
A particle moves in a circle of radius 0.5 m at 4 m/s. Its centripetal acceleration is:
8 m/s^2
16 m/s^2
32 m/s^2
2 m/s^2
Explanation: a_c = v^2/r = 16/0.5 = 32 m/s^2.
Question 16 of 20
The angular velocity of a body completing one revolution in 4 s (pi = 3.14) is:
0.79 rad/s
1.57 rad/s
3.14 rad/s
6.28 rad/s
Explanation: omega = 2 pi / T = 6.28 / 4 = 1.57 rad/s.
Question 17 of 20
The dot product of two perpendicular vectors is:
Maximum
AB
Zero
AB sin theta
Explanation: A . B = AB cos 90 = 0, since cos 90 = 0.
Question 18 of 20
Two angles of projection that give the same range are:
30 and 45 degrees
30 and 60 degrees
45 and 60 degrees
60 and 90 degrees
Explanation: Angles adding to 90 degrees give the same range; 30 + 60 = 90.
Question 19 of 20
An object thrown horizontally from a 20 m tower (g = 10 m/s^2) reaches the ground in:
1 s
2 s
4 s
0.5 s
Explanation: t = sqrt(2h/g) = sqrt(40/10) = sqrt(4) = 2 s.
Question 20 of 20
A wheel rotating at 120 rpm has a frequency of:
1 Hz
2 Hz
60 Hz
120 Hz
Explanation: f = 120/60 = 2 revolutions per second = 2 Hz.