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This is one of our many quizzes on Inelastic collisions. Click on this to access more such competitive quizzes on Mechanics.

Inelastic collisions Quiz 2

physics

A ball is dropped from a cliff of height $7\,m$, the coefficient of restitution between ball and ground is $0.85$. The maximum height reached by ball after the 2nd impact with the ground will be- 

 The formula for height after impact can easily be calculated by energy and momentum conservation of the system. 

 The maximum height reached after the first impact will be- $ \\  {{H}_{2}}={{e}^{2}}{{H}_{1}}\\ $ The maximum height reached after the 2nd impact will be- $ \\  {{H}_{3}}={{e}^{2}}{{H}_{2}}\\  \Rightarrow {{H}_{3}}={{e}^{2}}({{e}^{2}}{{H}_{1}})\\  \Rightarrow {{H}_{3}}={{e}^{4}}{{H}_{1}}\\  \Rightarrow {{H}_{3}}={{0.85}^{4}}\times 7\\  \therefore{{H}_{3}}=3.65\,m$

 With a velocity of $5\,m{{s}^{-1}}$, a moving ball collides with a stationary body. An observer seated on ball two notices that the relative velocity of both the body- 

The equal values of velocities can be calculated by energy and momentum conservation. 

  The velocities of both the balls after the collision will be equal as they are identical in nature, so for observer on ball two the relative velocity of another ball will remain zero in his frame of reference.

 A truck driver consumes alcohol, and loses control while driving, and collides with a car at rest in front of the truck at a velocity of $4\,m{{s}^{-1}}$. The total energy consumed in the process will be- $\text{[mass of car=500 kg, mass of truck=1500 kg]}$  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a8ff3c0c-0c78-4ac2-8cd5-42da2fd9e129308781126519551994.png"/>

The total momentum will remain conserved and the decrease in kinetic energy will result in converting to a different type of energy so kinetic energy will not remain conserved but total energy will be conserved. 

Applying momentum conservation $ \\  {{m}_{T}}{{v}_{1}}=({{m}_{T}}+{{m}_{C}})v\\  \Rightarrow 1500\times 4=(1500+500)v\\  \Rightarrow v=3m/s\\ $ Total energy consumed will be equal to change in kinetic energy of system so, $ \\  \Delta K.E=\dfrac{1}{2}{{m}_{T}}{{v}_{1}}^{2}-\dfrac{1}{2}({{m}_{T}}+{{m}_{C}}){{v}^{2}}\\  \Rightarrow \Delta K.E=\dfrac{1}{2}1500{{(4)}^{2}}-\dfrac{1}{2}(1500+500){{3}^{2}}\\  \therefore\Delta K.E=3000\,J$

 If a ball of mass $5\,kg$ is dropped from a height of 5 m and another identical ball of mass $3\,kg$ is dropped from a height of $7m$ the ratio of height achieved by both the balls after their 6th impact with the ground will be? $\left[\text{Here, e=0.7 }\right]$ 

During 6 impacts the height of the ball will decrease by a factor of ${{e}^{2}}$ in every impact. This relation can easily be calculated by momentum and energy conservation. 

 The height relation after and before the impact is given by $ \\  {{H}_{f}}={{e}^{2}}{{H}_{i}}\\ $ So after 6th impact ratio will be $ \\  \dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{{{e}^{12}}{{H}_{i}}}{{{e}^{12}}{{H}_{f}}}\\  \therefore\dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{5}{7}$

A police man fires a bullet of mass $30\,g$ with a speed of $5\,m{{s}^{-1}}$ towards dummy target of mass $700\,g$ in a practice section, after colliding with target bullet stick to it and both starts moving in the same direction, What will the change in kinetic energy of the system be? 

 The combined mass of bullet and target will provide momentum to the system and will remain conserved. 

 Momentum conservation $ \\  {{m}_{1}}{{v}_{1}}=({{m}_{1}}+{{m}_{2}})v\\  \Rightarrow 30\times 5=(730)v\\  \Rightarrow v=0.2m/s\\  \Rightarrow \Delta K.E=\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}-\dfrac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\\  \Rightarrow \Delta K.E=\dfrac{1}{2}0.3{{(5)}^{2}}-\dfrac{1}{2}(0.30+0.700){{(0.2)}^{2}}\\  \therefore \Delta K.E=3.69\,J$