20 NEET previous-year questions on Laws of Motion, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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9 ) g 1 ( g μ −
3 g 2 μ
3 ) 2 1 ( g μ −
2 ) 2 1 ( g μ −
Solution
For the system, the net force causing acceleration is the weight of minus the frictional forces on and . The equation of motion is . Thus, the correct option is (c).
24 Ns
20 Ns
12 Ns
6 Ns
Solution
The change in momentum is equal to the area under the force-time graph. The area from 0 to 8 s is the sum of the areas of the two triangles: . However, the second triangle is below the time axis, so its area is negative: . Thus, the correct option is (c).
a g ma 2 +
a g ma 2 −
a g ma +
a g ma −
Solution
For the balloon to ascend with the same acceleration , the effective weight must be reduced. The required mass to be removed is given by . Substituting and simplifying, the correct expression is , so option (a) is correct.
0.6 and 0.6
0.6 and 0.5
0.5 and 0.6
0.4 and 0.3
Solution
μs=tan30o= 1 √3 =0.5 μs=0.57=0.6 S= ut+1 2at2 4= 1 2a(4)2⇒a= 1 2 =0.5 www.vedantu.com 26 a=g sinθ−μ k(g)cosθ ⇒μK=0.9 √3 =0.5
2√2 3 v
3 4v
3 √2v
√3 2 v
Solution
www.vedantu.com 43 Pi⃗⃗ =Pf⃗⃗⃗ ⇒|Pi|=|Pf|⇒√(mV 3) 2 +(mV2)2 V2=2√2 3 V
(2t2 + 3t3)W
(2t2 + 4t4)W
(2t3 + 3t4)W
(2t3 + 3t5)W
Solution
a⃗ = 2tî + 3t2ĵ V⃗⃗ = 2t2î +3 3t3ĵ F⃗ = 2tî + 3t2ĵ P = F⃗ .V⃗⃗ = 2t3 + 3t5 www.vedantu.com 25
g, 3 g
3 g , g
g, g
3 g , 3 g
Solution
3m kx 3mg T Before the string is cut kx = T + 3 mg ...(1) T = mg ...(2) m T mg ⇒ kx = 4 mg After the string is cut, T = 0 a = 3 3 kx mg m - a = 43 3 mg mg m - 3m kx 3mg m mg ?↓ a = g a = 3 g ↑
0·8
0·25
0·5
0·4
Solution
Using the coefficient of restitution formula , where , , , and is the final velocity of the heavier block. By conservation of momentum, . Substituting, , so option (b) is correct.
Frictional force opposes the relative motion.
Limiting value of static friction is directly proportional to normal reaction.
Rolling friction is smaller than sliding friction.
Coefficient of sliding friction has dimensions of length.
Solution
The coefficient of sliding friction is a dimensionless quantity, as it is the ratio of the frictional force to the normal reaction. NCERT XI chapter Laws of Motion defines the coefficient of friction as a dimensionless constant, so option (d) is incorrect.
a = g cos θ
a = θ sin g
a = θ ec cos g
a = g tan θ
Solution
For the block to remain stationary on the wedge, the horizontal component of the normal force must balance the component of the gravitational force parallel to the incline. This gives , so option (d) is correct.
10 = rad/s
√10 = rad/s
10 2𝑣 rad/s
10 rad/s
Solution
𝑀𝑔= μ𝑁= μ𝑚ω2𝑟 ω = √ 𝑔 μ𝑟= √ 10 0.1×1=10 𝑟𝑎𝑑/𝑠
10%
( 3 13)%
16%
−10%
Solution
𝑑𝑥 𝑥 =2𝑑𝐴 𝐴 +1 2 𝑑𝐵 𝐵 +1 3 𝑑𝐶 𝐶 +3𝑑𝐷 𝐷 =2(1%)+1 2(2%)+1 3(3%)+3(4%) =2+1+1+12 =16%
Change according to the smallest force 𝑄𝑅→
Increase + + + + λ −λ P Q R 9 SPACE FOR ROUGH WORK
Decrease
Remain constant
Solution
Resultant force is zero.
g
g/2
g/5
g/10
Solution
For a system of two masses and connected by a massless string over a frictionless pulley, the acceleration is given by . Substituting and , , so option (c) is correct.
0 kg m/s
4.2 kg m/s
2.1 kg m/s
1.4 kg m/s
Solution
The impulse imparted to the ball is the change in momentum. The velocity just before and after impact can be found using . Substituting, . The change in momentum is , so option (b) is correct.
23000
20000
34500
23500
Solution
Fup = 2000g + 3000 = 23000 N Minimum power min ·= ρ ρP Fv min 323000 2= = ×P Fv = 34500 W
along south-west
along eastward
along northward
along north-east
Solution
The force required to change the direction of the player's motion is centripetal, acting perpendicular to the initial direction of motion. Since the player turns from southward to eastward, the force acts in the north-east direction, so option (d) is correct.
50 m s⁻²
1.2 m s⁻²
150 m s⁻²
1.5 m s⁻²
Solution
The maximum acceleration is given by , where is the coefficient of static friction. Substituting, , so option (d) is correct.
Zero
4 N
6 N
10 N
Solution
F = (M1 + M2)a –210 2 ms23a == + F′ = M2(2) = 3 × 2 N = 6 N
0.25
0.40
0.5
0.75
Solution
Squaring both sides
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