Collision — NEET Physics PYQ MCQs with Solutions

Solution: The smaller piece has mass $\dfrac{m}{6}$ and the larger $\dfrac{5m}{6}$. Momentum is conserved: $m\mathbf{v}=\dfrac{m}{6}\mathbf{v_1}+\dfrac{5m}{6}\mathbf{v_2}$, so $\mathbf{v}=\dfrac{\mathbf{v_1}}{6}+\dfrac{5\mathbf{v_2}}{6}$. With $\mathbf{v}=20\hat{i}+25\hat{j}-12\hat{k}$ and $\mathbf{v_1}=100\hat{i}+35\hat{j}+8\hat{k}$: $(120\hat{i}+150\hat{j}-72\hat{k})=(100\hat{i}+35\hat{j}+8\hat{k})+5\mathbf{v_2}$, giving $\mathbf{v_2}=\dfrac{1}{5}(20\hat{i}+115\hat{j}-80\hat{k})=4\hat{i}+23\hat{j}-16\hat{k}$.

Solution: The $5m$ particle splits into fragments of mass $m$, $m$ and $3m$. The two $m$ fragments move perpendicular to each other with speed $v$; the $3m$ fragment moves with $\mathbf{v'}$ so momentum is conserved: $5m\times0=mv\hat{i}+mv\hat{j}+3m\mathbf{v'}$, giving $\mathbf{v'}=-\dfrac{v}{3}\hat{i}-\dfrac{v}{3}\hat{j}$, so $|\mathbf{v'}|=\dfrac{v\sqrt{2}}{3}$. Energy released $E=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times3m\left(\dfrac{v\sqrt{2}}{3}\right)^2=mv^2+\dfrac{mv^2}{3}=\dfrac{4}{3}mv^2$.

Solution: In a head-on elastic collision, momentum and kinetic energy are conserved. Momentum: $4mu=4mv_1+2mv_2$, i.e. $2u=2v_1+v_2$ ...(i). Energy: $2mu^2=2mv_1^2+mv_2^2$, i.e. $2u^2=2v_1^2+v_2^2$ ...(ii). Solving, $v_1=\dfrac{1}{3}u$ and $v_2=\dfrac{4}{3}u$ (also $v_1=\dfrac{m_1-m_2}{m_1+m_2}u=\dfrac{4m-2m}{4m+2m}u=\dfrac{1}{3}u$). Decrease in KE of $A$: $\Delta KE=2m(u^2-v_1^2)=2m\left(u^2-\dfrac{u^2}{9}\right)=\dfrac{16mu^2}{9}$. Fraction lost $=\dfrac{\Delta KE}{KE_A}=\dfrac{16mu^2}{9}\times\dfrac{1}{2mu^2}=\dfrac{8}{9}$.

Solution: By conservation of linear momentum with $m_1=m$, $m_2=4m$, $u_1=v$, $u_2=0$, $v_1=0$: $mv=4mv_2$, so $v_2=\dfrac{v}{4}$ ...(i). Coefficient of restitution $e=\dfrac{v_2-v_1}{u_1-u_2}=\dfrac{\frac{v}{4}-0}{v-0}=\dfrac{1}{4}=0.25$.

Solution: In an elastic collision, kinetic energy of the system remains unchanged and momentum is also conserved. Since the balls have the same mass and the collision is perfectly elastic ($e=1$), their velocities are interchanged. Thus, $v'_A=v_B=-0.3$ m/s and $v'_B=v_A=0.5$ m/s.

Solution: By conservation of kinetic energy, $\dfrac{1}{2}Mv^2=\dfrac{1}{2}M\left(\dfrac{v}{3}\right)^2+\dfrac{1}{2}Mv_2^2$. So $v^2-\dfrac{v^2}{9}=v_2^2\Rightarrow v_2^2=\dfrac{8v^2}{9}$, giving the second block's speed $v_2=\dfrac{2\sqrt{2}}{3}v$.

Solution: The two $m$ pieces have perpendicular momenta of magnitude $mv$ each, giving a resultant $\sqrt{2}\,mv$. The third piece (mass $2m$) must carry equal and opposite momentum: $\sqrt{2}(mv)=(2m)v'$, so $v'=\dfrac{v}{\sqrt{2}}$. Total KE $=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv^2+\dfrac{1}{2}(2m)v'^2=mv^2+m\left(\dfrac{v}{\sqrt{2}}\right)^2=mv^2+\dfrac{mv^2}{2}=\dfrac{3}{2}mv^2$.

Solution: Using $v_1=\left[\dfrac{m_1-em_2}{m_1+m_2}\right]u_1+\left[\dfrac{(1+e)m_2}{m_1+m_2}\right]u_2$ with $u_1=2$, $u_2=0$, $m_1=m$, $m_2=2m$, $e=0.5$: $v_1=\left[\dfrac{m-m}{3m}\right]\times2=0$. Similarly $v_2=\left[\dfrac{(1+e)m_1}{m_1+m_2}\right]u_1=\left[\dfrac{1.5\times m}{3m}\right]\times2=1$ ms$^{-1}$.

Solution: Momentum of first part $=1\times12=12$ kg ms$^{-1}$; of second part $=2\times8=16$ kg ms$^{-1}$. As they are perpendicular, resultant momentum $=\sqrt{12^2+16^2}=20$ kg ms$^{-1}$. The third part carries equal and opposite momentum, so $4\times M=20\Rightarrow M=5$ kg.

Solution: Let the shell speed be $v$ and gun speed $V$. Conservation of momentum: $4V+0.2v=0$. Conservation of energy: $\dfrac{1}{2}\times4\times V^2+\dfrac{1}{2}\times0.2\times v^2=1050$. Solving the two equations gives $v=100$ m/s.

Solution: By conservation of momentum $0=m_1v_1-m_2v_2\Rightarrow\dfrac{v_1}{v_2}=\dfrac{m_2}{m_1}$. Ratio of kinetic energies $\dfrac{K_1}{K_2}=\dfrac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2}=\dfrac{m_1}{m_2}\times\left(\dfrac{m_2}{m_1}\right)^2=\dfrac{m_2}{m_1}$.

Solution: With $u_1=3$ m/s, $u_2=-5$ m/s and equal masses, momentum gives $v_1+v_2=-2$ ...(i). For an elastic collision $v_2-v_1=e(u_1-u_2)=(1)(3+5)=8\Rightarrow v_1-v_2=-8$ ...(ii). Adding, $2v_1=-10\Rightarrow v_1=-5$ m/s and $v_2=+3$ m/s. For equal masses in elastic collision the velocities are simply interchanged.

Solution: $v_1=36$ km/h $=10$ m/s, $v_2=0$, $m_1=2$ kg, $m_2=3$ kg. Common velocity $v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}=\dfrac{2\times10}{5}=4$ m/s. Loss in KE $=\dfrac{1}{2}m_1v_1^2-\dfrac{1}{2}(m_1+m_2)v^2=\dfrac{1}{2}\times2\times10^2-\dfrac{1}{2}\times5\times4^2=100-40=60$ J.

Solution: By conservation of linear momentum for this perfectly inelastic collision, $m\times3+2m\times0=(m+2m)v\Rightarrow 3m=3mv\Rightarrow v=1$ km/h.

Solution: When two bodies of equal masses undergo a head on elastic collision in one dimension, their velocities are simply interchanged, so $A$ takes $-0.3$ m/s and $B$ takes $+0.5$ m/s.

Solution: The coefficient of restitution $e=\dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}=\dfrac{v_2-v_1}{u_1-u_2}$. For a perfectly elastic collision the relative velocity of separation equals the relative velocity of approach, so $e=1$.