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NEET-UG and JEE Main Physics · Class 11

Momentum, Newton's second law and impulse — NEET & JEE Questions

Momentum, Newton's second law and impulse is a topic in Laws of Motion, a Class 11 Physics chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 18 published NEET-UG and JEE Main questions on Momentum, Newton's second law and impulse, each with the marking scheme of the real paper.

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Momentum, Newton's second law and impulse MCQs with solutions

10 of the 18 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    Linear momentum of a particle is defined as

    • A. p⃗=mv⃗\vec p=m\vec v
    • B. p⃗=ma⃗\vec p=m\vec a
    • C. p⃗=F⃗/m\vec p=\vec F/m
    • D. p⃗=m/v⃗\vec p=m/\vec v
  2. Question 2 · difficulty L2 · understanding

    A light unstretchable string passing over a smooth light pulley connects two blocks of masses m1m_1 and m2m_2. If the acceleration of the system is g8\frac{g}{8}, then the ratio of the masses m2m1\frac{m_2}{m_1} is :

    • A. 5:35: 3
    • B. 8:18: 1
    • C. 9:79: 7
    • D. 4:34: 3
  3. Question 3 · difficulty L2 · understanding

    A light string passing over a smooth light pulley connects two blocks of masses m1m_1 and m2(m_2\left(\right. where m2>m1)\left.m_2>m_1\right). If the acceleration of the system is g2\frac{g}{\sqrt{2}}, then the ratio of the masses m1m2\frac{m_1}{m_2} is:

    • A. 1+55−1\frac{1+\sqrt{5}}{\sqrt{5}-1}
    • B. 3+12−1\frac{\sqrt{3}+1}{\sqrt{2}-1}
    • C. 2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}
    • D. 1+52−1\frac{1+\sqrt{5}}{\sqrt{2}-1}
  4. Question 4 · difficulty L2 · understanding

    A light string passing over a smooth light fixed pulley connects two blocks of masses m1m_1 and m2m_2. If the acceleration of the system is g/8g / 8, then the ratio of masses is:

    • A. 81\frac{8}{1}
    • B. 97\frac{9}{7}
    • C. 53\frac{5}{3}
    • D. 43\frac{4}{3}
  5. Question 5 · difficulty L2 · understanding

    Two masses M1M_{1} and M2M_{2} are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2M_{2} is twice that of M1M_{1}, the acceleration of the system is a1a_{1}. When the mass M2M_{2} is thrice that of M1M_{1}, the acceleration of the system is a2a_{2}. The ratio a1a2\frac{a_{1}}{a_{2}} will be :

    • A. 13\frac{1}{3}
    • B. 23\frac{2}{3}
    • C. 32\frac{3}{2}
    • D. 12\frac{1}{2}
  6. Question 6 · difficulty L3 · understanding

    A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L = 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg-m 2 /s. The kinetic energy of the pendulum just after the lift-off is K Joules. The value of J is ___________.

  7. Question 7 · difficulty L3 · understanding

    A block of mass 5 kg5 \mathrm{~kg} moves along the xx-direction subject to the force F=(−20x+10)NF=(-20 x+10) \mathrm{N}, with the value of xx in metre. At time t=0 st=0 \mathrm{~s}, it is at rest at position x=1 mx=1 \mathrm{~m}. The position and momentum of the block at t=π4st={\pi \over 4} \mathrm{s} are

    • A. −0.5 m,5 kg m/s-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
    • B. 0.5 m,0 kg m/s0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
    • C. 0.5 m,−5 kg m/s0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
    • D. −1 m,5 kg m/s-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
  8. Question 8 · difficulty L3 · understanding

    Two balls, having linear momenta p→1=pi^{\overrightarrow p _1} = p\widehat i and p→2=−pi^{\overrightarrow p _2} = - p\widehat i, undergo a collision in free space. There is no external force acting on the balls. Let p′→1{\overrightarrow {p'} _1} and p′→2{\overrightarrow {p'} _2} be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of p,a1,a2,b1,b2,c1p,{a_1},{a_2},{b_1},{b_2},{c_1} and c2{c_2} :

    • A. p′→1=a1i^+b1j^+c1k^;p′→2=a2i^+b2j^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j
    • B. p′→1=c1k^;p′→2=c2k^{\overrightarrow {p'} _1} = {c_1}\widehat k;{\overrightarrow {p'} _2} = {c_2}\widehat k
    • C. p′→1=a1i^+b1j^+c1k^;p′→2=a2i^+b2j^−c1k^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j - {c_1}\widehat k
    • D. p′→1=a1i^+b1j^;p′→2=a2i^+b1j^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_1}\widehat j
  9. Question 9 · difficulty L3 · understanding

    In the given arrangement of a doubly inclined plane two blocks of masses MM and mm are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of mm, for which M=10 kgM=10 \mathrm{~kg} will move down with an acceleration of 2 m/s22 \mathrm{~m} / \mathrm{s}^2, is: (take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 and tan⁡37∘=3/4)\left.\tan 37^{\circ}=3 / 4\right)

    • A. 4.5 kg
    • B. 6.5 kg
    • C. 9 kg
    • D. 2.25 kg
  10. Question 10 · difficulty L3 · understanding

    As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s2^2)

    • A. (43−1)N\left( {4\sqrt 3 - 1} \right)N
    • B. (43+1)N\left( {4\sqrt 3 + 1} \right)N
    • C. 4(3−1)N4\left( {\sqrt 3 - 1} \right)N
    • D. 4(3+1)N4\left( {\sqrt 3 + 1} \right)N

Common mistakes in Momentum, Newton's second law and impulse

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Adds mass and speed instead of multiplying.
  • Divides force by time instead of multiplying.
  • Uses final momentum instead of change in momentum.
  • Assumes the same momentum change always requires the same force regardless of time.

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