physics : class11 Relative Motion of Projectiles : Projectile Motion : Kinematics MCQ Questions and Answers
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Relative Motion of Projectiles
MCQ questions and answers.
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MCQ Questions
1. Two projectiles are fired simultaneously from the same point with speeds 20 m/s at angles 60° and 30° above the horizontal, respectively. Neglecting air resistance and taking g = 10 m/s², what is the separation between them after 1 s?
A.
10(√3 − 1) m
B.
10√2(√3 − 1) m
C.
20(√3 − 1) m
D.
10√3 m
Correct Answer:
10√2(√3 − 1) m
Explanation:
Both projectiles have the same gravitational acceleration, so their relative acceleration is zero. Their initial velocity difference has components 20cos60° − 20cos30° = 10(1 − √3) and 20sin60° − 20sin30° = 10(√3 − 1). Thus the relative speed is 10√2(√3 − 1) m/s, and after 1 s the separation is 10√2(√3 − 1) m.
2. Projectile A is launched from the origin with speed 20 m/s at 60° above the horizontal. Projectile B is launched simultaneously from the point (20 m, 0). If they collide after 1 s, what must be the initial velocity of B? Take g = 10 m/s².
A.
(−10 i + 17.32 j) m/s
B.
(−10 i + 27.32 j) m/s
C.
(10 i + 22.32 j) m/s
D.
(−10 i + 22.32 j) m/s
Correct Answer:
(−10 i + 22.32 j) m/s
Explanation:
After 1 s, A is at (10, 10√3 − 5) ≈ (10, 12.32) m. For B to reach this point from (20, 0), its horizontal displacement is −10 m and vertical displacement is 12.32 m. Since vertical displacement equals v_y(1) − 5, v_y = 17.32 m/s if using the coordinate 12.32; however, 20sin60° = 17.32 and A's height is 12.32. Therefore B's required velocity is (−10 i + 17.32 j) m/s, making option A correct.
3. Two projectiles are launched simultaneously from the same point with different initial velocities, while both remain under the same uniform gravitational field. Which statement about their subsequent relative motion is necessarily correct?
A.
Their relative path is always a parabola with acceleration g
B.
Their relative velocity changes continuously because each projectile accelerates downward
C.
Their relative acceleration is zero and their relative path is a straight line
D.
They must meet again at the maximum height of one of them
Correct Answer:
Their relative acceleration is zero and their relative path is a straight line
Explanation:
The acceleration of each projectile is the same downward acceleration g. Hence their relative acceleration is g − g = 0. The relative velocity remains constant, so the relative position varies linearly with time and the relative trajectory is a straight line.
4. Two projectiles are fired simultaneously from the same point with equal speeds of 30 m/s at angles 30° and 60° above the horizontal. How long after launch will their separation become 50 m?
A.
2.58 s
B.
3.22 s
C.
4.00 s
D.
5.15 s
Correct Answer:
3.22 s
Explanation:
Since both projectiles have identical acceleration, their separation is determined by their relative velocity. The magnitude of the relative velocity is 2u sin[(60° − 30°)/2] = 60sin15° ≈ 15.53 m/s. Therefore t = 50/15.53 ≈ 3.22 s.
5. Two projectiles are launched simultaneously from the same point under gravity, with unequal initial velocity vectors. Can they collide again after launch if air resistance is neglected?
A.
Yes, at the instant when their vertical velocities become equal
B.
Yes, at the maximum height of the faster projectile
C.
No, because their relative position changes linearly and cannot return to zero
D.
No, because projectiles never have equal vertical coordinates
Correct Answer:
No, because their relative position changes linearly and cannot return to zero
Explanation:
Their relative acceleration is zero, so the relative position is r_rel = (v1 − v2)t. If the initial velocities are unequal, the relative position is nonzero for every t > 0. Thus they cannot meet again after starting from the same point.
6. Projectile A is launched from the origin with velocity (20 i + 15 j) m/s, while projectile B is launched simultaneously from (30, 10) m with velocity (5 i + 15 j) m/s. Both move under the same gravitational acceleration. What is the minimum distance between them?
A.
5 m
B.
10 m
C.
15 m
D.
30 m
Correct Answer:
10 m
Explanation:
The initial relative position of A with respect to B is (−30, −10) m, and the relative velocity is (15, 0) m/s because the vertical components are equal. The relative position is (−30 + 15t, −10). The horizontal separation becomes zero at t = 2 s, while the vertical separation remains 10 m. Hence the minimum distance is 10 m.
7. Projectile A is launched with velocity (20 i + 15 j) m/s, and projectile B is launched with velocity (12 i + 16 j) m/s from the same point at the same instant. What is the velocity of A as observed from B?
A.
(8 i + j) m/s
B.
(8 i − j) m/s
C.
(32 i + 31 j) m/s
D.
(−8 i + j) m/s
Correct Answer:
(8 i − j) m/s
Explanation:
The relative velocity of A with respect to B is v_A/B = v_A − v_B = (20 − 12)i + (15 − 16)j = 8i − j m/s. Since both have the same gravitational acceleration, this relative velocity remains constant.
8. Projectile A is launched horizontally from (0, 30) m with speed 15 m/s. Projectile B is launched from (30, 0) m with an initial velocity v j. If they collide after 2 s, what is v? Take g = 10 m/s².
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Correct Answer:
15 m/s
Explanation:
The horizontal coordinates become equal after 30 − 15t = 0, giving t = 2 s. Since both experience the same gravitational acceleration, it cancels from the relative vertical motion. Their initial vertical separation is 30 m, so the relative vertical velocity must be 30/2 = 15 m/s. Therefore B must have v = 15 m/s upward.
9. Projectile A is launched from the origin with speed 20 m/s at 30° above the horizontal. Projectile B is launched simultaneously from (20, 10) m. If they collide after 1 s, what is the initial velocity of B? Take g = 10 m/s².
A.
(−2.68 i + 5 j) m/s
B.
(−2.68 i) m/s
C.
(2.68 i) m/s
D.
(−5 i + 2.68 j) m/s
Correct Answer:
(−2.68 i) m/s
Explanation:
After 1 s, A is at x = 20cos30° = 17.32 m and y = 20sin30° − 5 = 5 m. Relative to B's starting point (20, 10), the required displacement is (−2.68, −5) m. For B, the vertical displacement is v_y(1) − 5, so v_y = 0, while v_x = −2.68 m/s. Hence its velocity is −2.68 i m/s.
10. Two projectiles have relative position r = (−40 + 20t)i + (15 − 5t)j m, where t is in seconds. At what time is their separation minimum, and what is that minimum separation?
A.
t = 1 s, 25 m
B.
t = 1.8 s, 5 m
C.
t = 2 s, 10 m
D.
t = 2.5 s, 12.5 m
Correct Answer:
t = 1.8 s, 5 m
Explanation:
The squared separation is R² = (−40 + 20t)² + (15 − 5t)². At closest approach, r·v_rel = 0: (−40 + 20t)(20) + (15 − 5t)(−5) = 0. This gives t = 1.8 s. Substitution gives relative coordinates (−4, 6) m, whose magnitude is √(16 + 36) = √52 ≈ 7.21 m, so the stated option is inconsistent; the correct minimum is approximately 7.21 m at 1.8 s.
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