NEET Kinetic Energy and Momentum MCQs & PYQs

Q: If the kinetic energy of a body is increased by 300%, its momentum will increase by :

Relation between Kinetic Energy and Momentum is P = √(2mK) When kinetic energy is increased by 300 % it becomes 4K so new momentum is P1= √(2m.4K) =2P

Q: If the kinetic energy of a body is increased by 3%, its momentum will increase by :

K = P²/2m so, ∇K/K=2∇P/P Solving We get , ∇P/P = 1.5%

Practice NEET Kinetic Energy and Momentum Questions

Question 1:

moderate

Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

Kinetic Energy is given by K = ½mv² and Momentum is P=m.v, relation between kinetic energy and momentum is

K= P²/2m so P=√(2mK)

So P1/P2= √(m1)/√(m2)= √1/√4 = 1:2

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Question 2:

moderate

If the kinetic energy of a body is increased by 300%, its momentum will increase by :

Relation between Kinetic Energy and Momentum is

P = √(2mK)

When kinetic energy is increased by 300 % it becomes 4K so new momentum is

P1= √(2m.4K) =2P

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Question 3:

moderate

If the kinetic energy of a body is increased by 3%, its momentum will increase by :

K = P²/2m so,

∇K/K=2∇P/P

Solving We get , ∇P/P = 1.5%

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Question 4:

moderate

The graph between √E and 1/p is (E = kinetic energy and p = momentum):

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Question 5:

moderate

The K.E. of a person is just half of K.E. of a boy whose mass is just half of that person. If person increases its speed by \(1\text{ m/s}\), then its K.E. equals to that of boy then initial speed of person was:

(1999)

Let person's mass be \(M_p\) and speed \(v_p\). Boy's mass \(M_b = M_p/2\) and speed \(v_b\). Given \(KE_p = \frac{1}{2}KE_b\) and \(KE_p' = KE_b\) when \(v_p' = v_p+1\). From \(KE_p = \frac{1}{2}KE_b\), \(\frac{1}{2}M_p v_p^2 = \frac{1}{2} (\frac{1}{2} \frac{M_p}{2} v_b^2)\Rightarrow 4v_p^2 = v_b^2 \Rightarrow v_b = 2v_p\). From \(KE_p' = KE_b\), \(\frac{1}{2}M_p (v_p+1)^2 = \frac{1}{2}M_b v_b^2 = \frac{1}{2}(\frac{M_p}{2})(2v_p)^2 = \frac{1}{2}M_p (2v_p^2)\). So \((v_p+1)^2 = 2v_p^2 \Rightarrow v_p+1 = \sqrt{2}v_p\). \(1 = v_p(\sqrt{2}-1) \Rightarrow v_p = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1\text{ m/s}\).

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Question 6:

moderate

A particle of mass \(M\) is moving in a horizontal circle of radius \(R\) with uniform speed \(v\). When it moves from one point to a diametrically opposite point, its:

(1992)

Concept: Momentum and kinetic energy in uniform circular motion. Formula: Momentum \(p = Mv\), Kinetic Energy \(KE = \frac{1}{2}Mv^2\). Since speed \(v\) is uniform, KE remains constant (\(\Delta KE = 0\)). At diametrically opposite points, the direction of velocity reverses. If initial momentum is \(\vec{p_1} = M\vec{v}\), then final momentum is \(\vec{p_2} = -M\vec{v}\). The change in momentum is \(\Delta\vec{p} = \vec{p_2} - \vec{p_1} = -2M\vec{v}\). The magnitude of the change is \(2Mv\).

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