Motion in a Plane — Practice

Question 1

A vector has components Ax = 3 and Ay = 4. Find its magnitude and direction.

Accepted Answer

Magnitude A = sqrt(3^2 + 4^2) = sqrt(9 + 16) = 5. Direction theta = tan^-1(4/3) = 53.1 degrees measured from the x-axis. Both components are positive, so the vector lies in the first quadrant and the inverse tangent needs no correction.

Question 2

Evaluate i.j and i x j for the unit vectors along the x and y axes.

Accepted Answer

i.j = 0, because the two unit vectors are perpendicular and the scalar product carries cos 90 = 0. i x j = k, a unit vector along z, because the vector product is largest for perpendicular vectors and the right-hand rule fixes the direction. Note how one vanishes exactly where the other is maximum.

Question 3

A stone is thrown horizontally from a cliff. What is its vertical velocity the instant it leaves the hand, and how does that affect the time it takes to land?

Accepted Answer

The initial vertical velocity is zero, because the throw is purely horizontal. So the vertical motion is a plain drop from rest and the time to land is exactly the same as for a stone simply dropped from the same height. The horizontal speed changes where it lands, never when.

Question 4

Two forces of 6 N and 8 N act at a point with 60 degrees between them. Find the magnitude of the resultant.

Accepted Answer

R = sqrt(A^2 + B^2 + 2AB cos theta) = sqrt(36 + 64 + 2 x 6 x 8 x cos 60) = sqrt(36 + 64 + 96 x 0.5) = sqrt(100 + 48) = sqrt(148) = 12.2 N. Note it is less than 6 + 8 = 14 N; the vectors only add arithmetically when they are parallel.

Question 5

A projectile is fired at 50 m/s at 30 degrees to the horizontal. Find the time of flight, maximum height and range (g = 10 m/s^2).

Accepted Answer

Vertical component uy = 50 sin 30 = 25 m/s. Time of flight T = 2uy/g = 50/10 = 5 s. Maximum height H = u^2 sin^2(theta)/2g = 2500 x 0.25/20 = 31.25 m. Range R = u^2 sin(60)/g = 2500 x 0.866/10 = 216.5 m.

Question 6

A stone is thrown horizontally at 20 m/s from a height of 80 m. Find its horizontal range (g = 10 m/s^2).

Accepted Answer

The vertical motion is a drop from rest: 80 = (1/2)(10)t^2, giving t^2 = 16 and t = 4 s. The horizontal velocity is unchanged throughout, so the range is 20 x 4 = 80 m.

Question 7

A particle moves in a circle of radius 20 cm at 5 revolutions per second. Find its linear speed and centripetal acceleration.

Accepted Answer

Angular speed omega = 2 pi f = 10 pi rad/s. Linear speed v = omega R = 10 pi x 0.2 = 2 pi = 6.28 m/s. Centripetal acceleration a_c = omega^2 R = (10 pi)^2 x 0.2 = 20 pi^2 = 197 m/s^2, directed toward the centre at every instant.

Question 8

For A = 3i + 4j and B = 4i - 3j, find A.B and comment on the angle between them.

Accepted Answer

A.B = (3)(4) + (4)(-3) = 12 - 12 = 0. A zero scalar product with both vectors non-zero means cos(theta) = 0, so the vectors are perpendicular. Both have magnitude 5, so this is a pair of equal-length vectors at right angles.

Question 9

A ball is thrown at 20 m/s at 60 degrees to the horizontal. Find its speed and the direction of its velocity 1.0 s after launch (g = 10 m/s^2).

Accepted Answer

Components at launch: ux = 20 cos 60 = 10 m/s, uy = 20 sin 60 = 17.3 m/s. After 1.0 s the horizontal component is unchanged at 10 m/s, while the vertical is vy = 17.3 - 10(1.0) = 7.3 m/s. Speed = sqrt(10^2 + 7.3^2) = sqrt(100 + 53.3) = 12.4 m/s. Direction = tan^-1(7.3/10) = 36.1 degrees above the horizontal, so the ball is still rising but has flattened out from its launch angle of 60 degrees.

Question 10

Show that for a given launch speed, the ranges at 25 degrees and at 65 degrees are equal, and state which of the two reaches the greater height.

Accepted Answer

Range depends on the angle only through sin(2 theta). At 25 degrees, sin(50) = 0.766; at 65 degrees, sin(130) = sin(180 - 130) = sin(50) = 0.766. The two are identical, so the ranges are equal — the angles are equally spaced either side of 45 degrees. Maximum height goes as sin^2(theta), which increases all the way to 90 degrees, so the 65-degree throw rises much higher: sin^2(65) = 0.82 against sin^2(25) = 0.18, more than four times the height for the same distance.

Practice — Motion in a Plane

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