Centre of Mass and Momentum Practice Questions
JEE · Physics · 1408 free MCQs with instant results and detailed explanations.
1408
Total
281
Easy
604
Medium
523
Hard
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Topics in Centre of Mass and Momentum
Collisions in 1D and 2D
333
Impulse
243
Variable Mass Systems
246
Linear Momentum
297
Centre of Mass
289
Sample Questions from Centre of Mass and Momentum
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
What is the center of mass of a uniform rod of length L?
A.
L/2
B.
L/4
C.
L/3
D.
0
Show Answer & Explanation
Correct Answer: A
For a uniform rod, the center of mass is located at its midpoint, which is at a distance L/2 from either end.
Q2
Easy
If two particles of masses m1 and m2 are located at positions x1 and x2, respectively, what is the position of the center of mass (COM)?
A.
(m1*x1 + m2*x2)/(m1 + m2)
B.
(x1 + x2)/2
C.
(m1 + m2)/(x1 + x2)
D.
m1*x1 + m2*x2
Show Answer & Explanation
Correct Answer: A
The center of mass is calculated as the weighted average of the positions, which is given by (m1*x1 + m2*x2)/(m1 + m2).
Q3
Easy
A system consists of two blocks of masses 2 kg and 3 kg placed at distances of 4 m and 2 m from a reference point. What is the position of the center of mass?
A.
2.4 m
B.
3.6 m
C.
4.0 m
D.
5.0 m
Show Answer & Explanation
Correct Answer: B
Using the formula for COM, we find the weighted average: (2*4 + 3*2)/(2 + 3) = 3.6 m.
Q4
Medium
Two particles of masses 2 kg and 3 kg are located at points (2, 3) and (4, 5) respectively on a Cartesian plane. What is the position of the center of mass of the system?
A.
(3.6, 4.2)
B.
(3.0, 4.0)
C.
(3.2, 4.4)
D.
(2.5, 3.5)
Show Answer & Explanation
Correct Answer: A
The center of mass can be calculated using the formula: \( R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \). Here, \( R_{cm} = \frac{(2 \cdot (2,3) + 3 \cdot (4,5))}{2 + 3} = (3.6, 4.2) \).
Q5
Medium
A uniform rod of length 2 m and mass 10 kg is pivoted at one end and allowed to rotate in a vertical plane. What is the distance from the pivot to the center of mass of the rod?
A.
0.5 m
B.
1.0 m
C.
0.75 m
D.
1.5 m
Show Answer & Explanation
Correct Answer: B
For a uniform rod, the center of mass lies at its midpoint. Therefore, the distance from the pivot to the center of mass is half the length of the rod, which is 1.0 m.
Q6
Medium
In a system with three particles of masses 1 kg, 2 kg, and 3 kg located at (1, 0), (0, 2), and (2, 2) respectively, which of the following points is closest to the center of mass of the system?
A.
(1.2, 1.2)
B.
(1, 1)
C.
(2, 2)
D.
(1.5, 1.5)
Show Answer & Explanation
Correct Answer: A
The center of mass is given by \( R_{cm} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3}{m_1 + m_2 + m_3} \). Calculating gives us the center of mass at (1.2, 1.2).
Q7
Medium
Two point masses, m1 = 2 kg and m2 = 3 kg, are located at coordinates (1, 2) m and (4, 6) m respectively. What is the coordinates of the center of mass?
A.
(2.8, 4.2)
B.
(3.0, 4.0)
C.
(2.5, 3.5)
D.
(3.5, 5.0)
Show Answer & Explanation
Correct Answer: A
The center of mass (CM) can be calculated using the formula: CM_x = (m1*x1 + m2*x2) / (m1 + m2), CM_y = (m1*y1 + m2*y2) / (m1 + m2). After substituting the values, we find the CM coordinates.
Q8
Hard
A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod just before it strikes the horizontal surface?
A.
ā(2g/L)
B.
ā(g/L)
C.
ā(3g/L)
D.
ā(4g/L)
Show Answer & Explanation
Correct Answer: A
Using conservation of energy: potential energy at the top (MgL/2) converts to rotational kinetic energy (1/2 I Ļ^2). The moment of inertia I for a rod about one end is (1/3 ML^2). Setting the energies equal gives Ļ = ā(3g/L).
Q9
Hard
A projectile is launched with an initial velocity of 20 m/s at an angle of 30 degrees to the horizontal. Calculate the horizontal distance traveled by the projectile before it hits the ground.
A.
40 m
B.
60 m
C.
80 m
D.
100 m
Show Answer & Explanation
Correct Answer: B
The range R of a projectile is given by R = (v^2 * sin(2Īø))/g. Plugging in v = 20 m/s, Īø = 30 degrees, and g = 9.81 m/sĀ² gives R = 60 m.
Q10
Hard
A uniform rod of length L and mass M is pivoted at one end. What is the position of its center of mass?
A.
L/2
B.
L/3
C.
L/4
D.
L/6
Show Answer & Explanation
Correct Answer: B
For a uniform rod, the center of mass is located at L/3 from the pivot towards the free end. This is derived from the definition of center of mass for continuous bodies.
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Centre of Mass and Momentum ā JEE Physics Practice Questions Online
This page contains 1408 practice MCQs for the chapter Centre of Mass and Momentum in JEE Physics. The questions are organized by difficulty ā 281 easy, 604 medium, 523 hard ā so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit. This chapter covers 5 topics, giving you comprehensive coverage of the entire chapter.