Q & A (Physics)

Q: A body of mass m at rest is subjected to a constant force F for time t. The kinetic energy at time t is given by

(a) F²t² /2m

(b) F²t² /3m

(c) 2 F²t²/m

(d) F²t²/m

Ans: (a)

Solution: Acceleration , a = F/m

Applying , v = u + a t

$\displaystyle v = 0 + \frac{F}{m} t $

$\displaystyle v =  \frac{F}{m} t $

Kinetic Energy , $\displaystyle K = \frac{1}{2} m v^2  $

$\displaystyle K = \frac{1}{2} m (\frac{F}{m} t)^2  $

$\displaystyle K = \frac{F^2 t^2}{2 m}  $

Q: Two bodies of masses m and 4m have equal kinetic energy. What is the ratio of their momentum?

(a)1 : 4

(b) 1 : 2

(c)1 : 1

(d) 2 : 1

Ans: (b)

Sol: As Kinetic Energy $\displaystyle K = \frac{p^2}{2 m}$ ; p = Linear Momentum

K.E₁ = K.E₂

$\displaystyle \frac{p_1^2}{2 m} = \frac{p_2^2}{2 \times 4 m}$

$\displaystyle \frac{p_1^2}{p_2^2} = \frac{1}{4}$

$\displaystyle \frac{p_1}{p_2} = \frac{1}{2}$

Q: An electron, a proton, a deutron and an  α-particle have same linear momenta. Which one of them possesses least kinetic energy ?

(a) Electron

(b) Proton

(c) Deutron

(d)  α-particle

Ans: (d)

Sol: The relation between K.E & Linear Momentum is

$\displaystyle K = \frac{p^2}{2 m}$ ; p = Linear momentum

$\displaystyle K \propto \frac{1}{m}$ ; As p = constant

Since α-particle has larger mass , hence it has least kinetic energy .

Q: A gun of mass M fires a bullet of mass m with maximum speed v. Given that m < M .The kinetic energy of the gun will be

(a) 1/2 mv²

(b) 1/2 Mv²

(c) more than 1/2 mv²

(d) less than 1/2 mv²

Ans: (d)

Sol: $\displaystyle 0 = m \vec{v} + M \vec{v_1}$

$ v_1 = – \frac{m}{M} v $

Kinetic energy of the gun will be

$ K.E = \frac{1}{2} M v_1^2 $

$ K.E = \frac{1}{2} M (\frac{m}{M} v)^2 $

$ K.E = \frac{1}{2 M} m^2 v^2 $

$ K.E = \frac{m}{M} ( \frac{1}{2 } m v^2 ) < \frac{1}{2 } m v^2 $

Q:  Two bodies A and B and having masses m and M respectively possess same momenta. Given that M > m . If E_A and E_B be their kinetic energies , then which of the following statement is true ?

(a) E_A = E_B

(b) E_A <  E_B

(c) E_A > E_B

(d) It cannot be predicted.

Ans: (c)

$\displaystyle E_A = \frac{p^2}{2 m}$

$\displaystyle E_B = \frac{p^2}{2 M}$

$\displaystyle  \frac{E_A}{E_B} = \frac{M}{m} > 1 $

E_A > E_B