2-2 Motion of connected bodies: practice questions
Applied Mathematics — Second: Dynamics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 37 questions on this lesson, each with an explanation.
A body hangs from a spring balance fixed to the ceiling of a lift. The balance reads 13 kg.wt when the lift moves up with a uniform acceleration and 7 kg.wt when it moves down with the same acceleration. Find the real weight of the body.
A10 kg.wt
B20 kg.wt
C3 kg.wt
D13 kg.wt
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Correct answer: (A) 10 kg.wt
Explanation: If the lift moves up with a uniform acceleration and down with the same acceleration, the reading going up + the reading going down = twice the real weight.
Real weight = (13 + 7) ÷ 2 = 10 kg.wt.
Question 2Medium
Two masses of 5 kg and 3 kg hang from the ends of a light string passing over a smooth fixed pulley. Find the acceleration of the system. (Take g = 9.8 m/sec².)
A1.225 m/sec²
B12.25 m/sec²
C19.6 m/sec²
D2.45 m/sec²
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Correct answer: (D) 2.45 m/sec²
Explanation: Step 1: Let the acceleration be a and the tension T. For the 5 kg (moving down): 5g − T = 5a. For the 3 kg (moving up): T − 3g = 3a.
Step 2: Add the two equations: 5g − 3g = 8a, so 2g = 8a.
Step 3: a = (5 − 3) g ÷ (5 + 3) = 2 × 9.8 ÷ 8 = 2.45.
∴ Acceleration = 2.45 m/sec²
Question 3Medium
A man of mass 75 kg stands on the floor of a lift moving downwards with a uniform retardation of 0.98 m/sec² (its speed is decreasing). Find the pressure of the man on the floor of the lift (g = 9.8 m/sec²).
A67.5 kg.wt
B75 kg.wt
C73.5 kg.wt
D82.5 kg.wt
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Correct answer: (D) 82.5 kg.wt
Explanation: Step 1: The lift moves down but slows down, so its acceleration is 0.98 m/sec² upwards.
Step 2: Taking upwards as positive: m a = R − m g, so R = m (g + a) = 75 (9.8 + 0.98) = 75 × 10.78 = 808.5 N.
Step 3: R = 808.5 ÷ 9.8 = 82.5 kg.wt, and the pressure on the floor equals the reaction: 82.5 kg.wt (67.5 kg.wt would be for a lift speeding up downwards).
Question 4Medium
A lift of mass 500 kg is carried by a cable that cannot bear a tension greater than 1500 kg.wt. Find the greatest load that can be placed in the lift when it moves upwards with a uniform acceleration of 2.45 m/sec² (g = 9.8 m/sec²).
A1000 kg
B1200 kg
C500 kg
D700 kg
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Correct answer: (D) 700 kg
Explanation: Step 1: Let the load be m kg. For the lift and the load together: (500 + m) a = T − (500 + m) g.
Step 2: With T = 1500 × 9.8 N: (500 + m)(2.45 + 9.8) = 1500 × 9.8.
Step 3: 500 + m = 14700 ÷ 12.25 = 1200.
Step 4: m = 700 kg.
Question 5Medium
A body of mass 50 kg stands on the floor of a lift that accelerates downward at 2 m/sec². Find the reaction of the floor on the body. (Take g = 9.8 m/sec².)
A390 N
B100 N
C490 N
D590 N
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Correct answer: (A) 390 N
Explanation: Step 1: Forces on the body: reaction R upward, weight m g downward. The acceleration is downward.
Step 2: Apply F = m a downward: m g − R = m a.
Step 3: R = m(g − a) = 50 × (9.8 − 2) = 50 × 7.8 = 390.
∴ Reaction = 390 N
Question 6Hard
A lift moves upward with acceleration 2 m/sec². A body of 2 kg hangs from the ceiling by a light string, and a body of 3 kg hangs from the 2 kg body by a second light string. Find the tension in the upper string. (g = 9.8 m/sec²)
A49 N
B35.4 N
C59 N
D69 N
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Correct answer: (C) 59 N
Explanation: Step 1: The upper string supports both bodies (5 kg) moving with a = 2 m/sec² upward: T − 5g = 5a.
Step 2: T = 5(9.8 + 2) = 5 × 11.8.
∴ T = 59 N; 35.4 N is the tension in the lower string.
Question 7Hard
Two bodies of masses 300 gm and 500 gm are attached to the ends of a light string passing over a small smooth pulley. The system starts from rest with the two bodies in the same horizontal plane, and the string is cut 2 seconds after the start. Find the vertical distance between the two bodies 1 second after the string is cut, given that neither reaches the ground. (g = 9.8 m/sec²)
A9.8 m
B14.7 m
C19.6 m
D29.4 m
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Correct answer: (C) 19.6 m
Explanation: a = (500 − 300) g ÷ 800 = 2.45 m/sec². After 2 s: v = 4.9 m/sec and each body has moved ½ × 2.45 × 4 = 4.9 m, so they are 9.8 m apart.
After the cut both move with acceleration g downwards, one moving down and the other up at 4.9 m/sec, so the distance grows at the constant rate 2 × 4.9 = 9.8 m/sec.
Distance = 9.8 + 9.8 = 19.6 m. (9.8 m forgets the gap at the moment of cutting.)
Question 8Hard
A man of mass 60 kg stands on a weighing machine in a lift that starts from rest and moves upward. The machine reads 66 kg.wt for the first 2 seconds, then 60 kg.wt for the next 6 seconds, then 54 kg.wt until the lift stops. Find the total distance moved by the lift. (g = 9.8 m/sec²)
A13.72 m
B1.6 m
C17.64 m
D15.68 m
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Correct answer: (D) 15.68 m
Explanation: Step 1: R − mg = ma: (66 − 60) × 9.8 = 60a ⇒ a = 0.98 m/sec² for 2 s, so v = 1.96 m/sec and s₁ = ½ × 0.98 × 2² = 1.96 m; then s₂ = 1.96 × 6 = 11.76 m.
Step 2: In the last stage (54 − 60) × 9.8 = 60a ⇒ a = −0.98 m/sec², so the lift stops after 1.96 ÷ 0.98 = 2 s and moves s₃ = ½ × 1.96 × 2 = 1.96 m.
∴ Total = 1.96 + 11.76 + 1.96 = 15.68 m; forgetting to convert kg.wt into newtons (× 9.8) gives the wrong 1.6 m.
Question 9Hard
Two scale pans, each of mass 50 gm, hang from the ends of a light string passing over a small smooth pulley. A body of mass 250 gm is put in one pan and a body of mass 50 gm in the other, and the system moves from rest. Find the pressure of the 250 gm body on its pan during the motion.
A250 gm.wt
B150 gm.wt
C125 gm.wt
D375 gm.wt
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Correct answer: (C) 125 gm.wt
Explanation: The sides are 300 gm and 100 gm: 400 a = 200 × 980 ⇒ a = 490 cm/sec², the heavier side moving down.
For the 250 gm body alone (moving down): 250 a = 250 × 980 − R ⇒ R = 250 × 490 dyne = 125 gm.wt, and its pressure on the pan equals R.
(150 gm.wt is the tension in the string; 250 gm.wt is the real weight.)
Question 10Hard
A lift of total mass 200 kg is raised by a cable with an upward acceleration of 1.5 m/sec². Find the tension in the cable. (Take g = 9.8 m/sec².)
A1960 N
B1660 N
C2260 N
D300 N
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Correct answer: (C) 2260 N
Explanation: Step 1: Forces on the lift: tension T up, weight mg down; acceleration is upward.
Step 2: T − m g = m a, so T = m(g + a).
Step 3: T = 200 × (9.8 + 1.5) = 200 × 11.3 = 2260.
∴ Tension = 2260 N