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Question: Masses and velocities at an instant of blocks $A$ and $B$ connected by an ideal spring (force consta...

Masses and velocities at an instant of blocks AA and BB connected by an ideal spring (force constant kk) are shown. The elongation of the spring at the shown instant is half of the subsequent maximum elongation. Then in the subsequent motion

A

The minimum kinetic energy of the system of AA and BB is 25mv022\frac{25 mv_0^2}{2}

B

The maximum kinetic energy of the system of AA and BB is 29mv022\frac{29 mv_0^2}{2}

C

The maximum elongation of the spring is v0mkv_0 \sqrt{\frac{m}{k}}

D

The maximum acceleration of block AA is 2v0km2v_0 \sqrt{\frac{k}{m}}

Answer

A, B, D

Explanation

Solution

The system consists of two blocks connected by a spring on a smooth surface. The total momentum of the system is conserved since there are no external horizontal forces. The velocity of the center of mass is constant.

At the given instant, mass of block A is mm, velocity is vA=4v0v_A = 4v_0. Mass of block B is 3m3m, velocity is vB=2v0v_B = 2v_0.

Total momentum P=mvA+3mvB=m(4v0)+3m(2v0)=4mv0+6mv0=10mv0P = mv_A + 3mv_B = m(4v_0) + 3m(2v_0) = 4mv_0 + 6mv_0 = 10mv_0.

Velocity of center of mass vCM=Pm+3m=10mv04m=52v0v_{CM} = \frac{P}{m+3m} = \frac{10mv_0}{4m} = \frac{5}{2}v_0.

The total energy of the system is conserved: E=K+UE = K + U, where KK is the kinetic energy and UU is the potential energy of the spring.

K=12mvA2+12(3m)vB2K = \frac{1}{2}mv_A^2 + \frac{1}{2}(3m)v_B^2. U=12kx2U = \frac{1}{2}kx^2, where xx is the elongation of the spring.

Minimum kinetic energy occurs when the relative velocity of the blocks is zero, which means vA=vB=vCMv_A = v_B = v_{CM}.

Kmin=12mvCM2+12(3m)vCM2=12(m+3m)vCM2=12(4m)(52v0)2=2m254v02=252mv02K_{min} = \frac{1}{2}m v_{CM}^2 + \frac{1}{2}(3m) v_{CM}^2 = \frac{1}{2}(m+3m)v_{CM}^2 = \frac{1}{2}(4m)(\frac{5}{2}v_0)^2 = 2m \frac{25}{4}v_0^2 = \frac{25}{2}mv_0^2.

At minimum kinetic energy, the potential energy of the spring is maximum, Umax=12kxmax2U_{max} = \frac{1}{2}kx_{max}^2, where xmaxx_{max} is the maximum elongation.

Maximum kinetic energy occurs when the potential energy of the spring is minimum, which is Umin=0U_{min} = 0 when the spring is at its natural length.

Kmax=EUmin=EK_{max} = E - U_{min} = E. Kmin=EUmaxK_{min} = E - U_{max}.

At the initial instant, Ki=12m(4v0)2+12(3m)(2v0)2=12m(16v02)+12(3m)(4v02)=8mv02+6mv02=14mv02K_i = \frac{1}{2}m(4v_0)^2 + \frac{1}{2}(3m)(2v_0)^2 = \frac{1}{2}m(16v_0^2) + \frac{1}{2}(3m)(4v_0^2) = 8mv_0^2 + 6mv_0^2 = 14mv_0^2.

Let the elongation at the initial instant be xix_i. We are given xi=12xmaxx_i = \frac{1}{2}x_{max}.

Ui=12kxi2=12k(12xmax)2=18kxmax2U_i = \frac{1}{2}kx_i^2 = \frac{1}{2}k(\frac{1}{2}x_{max})^2 = \frac{1}{8}kx_{max}^2.

Total energy E=Ki+Ui=14mv02+18kxmax2E = K_i + U_i = 14mv_0^2 + \frac{1}{8}kx_{max}^2.

Also, E=Kmin+Umax=252mv02+12kxmax2E = K_{min} + U_{max} = \frac{25}{2}mv_0^2 + \frac{1}{2}kx_{max}^2.

Equating the two expressions for E:

14mv02+18kxmax2=252mv02+12kxmax214mv_0^2 + \frac{1}{8}kx_{max}^2 = \frac{25}{2}mv_0^2 + \frac{1}{2}kx_{max}^2.

14mv02252mv02=12kxmax218kxmax214mv_0^2 - \frac{25}{2}mv_0^2 = \frac{1}{2}kx_{max}^2 - \frac{1}{8}kx_{max}^2.

28252mv02=(418)kxmax2\frac{28-25}{2}mv_0^2 = (\frac{4-1}{8})kx_{max}^2.

32mv02=38kxmax2\frac{3}{2}mv_0^2 = \frac{3}{8}kx_{max}^2.

mv02=14kxmax2mv_0^2 = \frac{1}{4}kx_{max}^2.

xmax2=4mv02kx_{max}^2 = \frac{4mv_0^2}{k}.

xmax=2v0mkx_{max} = 2v_0 \sqrt{\frac{m}{k}}.

Now let's evaluate the options:

(A) The minimum kinetic energy of the system of A and B is 25mv022\frac{25 mv_0^2}{2}. This is correct, as calculated above.

(B) The maximum kinetic energy of the system of A and B is Kmax=EUmin=E0=14mv02+18kxmax2K_{max} = E - U_{min} = E - 0 = 14mv_0^2 + \frac{1}{8}kx_{max}^2.

Substitute xmax2=4mv02kx_{max}^2 = \frac{4mv_0^2}{k}:

Kmax=14mv02+18k(4mv02k)=14mv02+48mv02=14mv02+12mv02=(14+12)mv02=28+12mv02=292mv02K_{max} = 14mv_0^2 + \frac{1}{8}k(\frac{4mv_0^2}{k}) = 14mv_0^2 + \frac{4}{8}mv_0^2 = 14mv_0^2 + \frac{1}{2}mv_0^2 = (14 + \frac{1}{2})mv_0^2 = \frac{28+1}{2}mv_0^2 = \frac{29}{2}mv_0^2.

So, option (B) is correct.

(C) The maximum elongation of the spring is v0mkv_0 \sqrt{\frac{m}{k}}. Our calculation gives xmax=2v0mkx_{max} = 2v_0 \sqrt{\frac{m}{k}}. So, option (C) is incorrect.

(D) The maximum acceleration of block A is 2v0km2v_0 \sqrt{\frac{k}{m}}. The force on block A is FA=kxF_A = -kx, so the acceleration is aA=kmxa_A = -\frac{k}{m}x. The maximum magnitude of acceleration occurs at maximum elongation or compression, i.e., x=xmax|x| = x_{max}.

aAmax=kmxmax=km(2v0mk)=2v0k2m2mk=2v0km|a_A|_{max} = \frac{k}{m}x_{max} = \frac{k}{m} (2v_0 \sqrt{\frac{m}{k}}) = 2v_0 \sqrt{\frac{k^2}{m^2} \frac{m}{k}} = 2v_0 \sqrt{\frac{k}{m}}.

So, option (D) is correct.

Therefore, options (A), (B), and (D) are correct. Since this is a multiple choice question, there might be multiple correct options.