Work,Energy and Power Quiz-15
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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Q1. Two spherical bodies of the same mass M are moving with velocities v1 and v2.These collide perfectly inelastically , then the loss in kinetic energy is?
Solution
Loss of kinetic energy =1/2 (m1 m2)/(m1+m2 ) (v1-v2 )2 =1/2×(M×M)/(M+M) (V1-V2 )2 =(M∙M)/2(2M) (V1-V2 )2 =M/4(V1-V2)
Loss of kinetic energy =1/2 (m1 m2)/(m1+m2 ) (v1-v2 )2 =1/2×(M×M)/(M+M) (V1-V2 )2 =(M∙M)/2(2M) (V1-V2 )2 =M/4(V1-V2)
Q2. A position-dependent force F=3x2-2x+7 acts on a body of mass 7 kg and displaces it from x=0 m to x=5m. The work done on the body is x' joule. If both F and x are measured in SI units, the value of x' is?
Solution
This is the case of work done by a variable force
or W=(5×5×5-5×5+7×5) or W=(125-25+35)=135 J
This is the case of work done by a variable force
or W=(5×5×5-5×5+7×5) or W=(125-25+35)=135 J
Q3. A bullet is fired from a rifle. If the riffle recoils freely, then the kinetic energy of the rifle is?
Solution
E= p2/2m. If P= constant then E∝1/m i.e., kinetic energy of heavier body will be less. As the mass of gun is more than bullet therefore it possess less kinetic energy
E= p2/2m. If P= constant then E∝1/m i.e., kinetic energy of heavier body will be less. As the mass of gun is more than bullet therefore it possess less kinetic energy
Q4. In an inelastic collision, what is conserved?
Solution
Momentum
Momentum
Q5. A 2.0 kg block is dropped from a height of 40 cm onto a spring of spring constant k=1960 Nm-1. Find the maximum distance the spring is compressed?
Solution
Let m be the mass of the block, h the height from which it is dropped, and x the compression o the spring. Since, energy is conserved, so Final gravitational potential energy = final spring potential energy or mg(h+x)=1/2 kx2 or mg(h+x)+1/2 kx2=0 or kx2-2mg(h+x)=0 kx2-2mgx-2mgh=0 This is a quadratic equation for x. Its solution is x=(mg± √((mg)2+2mghk))/k Now, mg=2×9.8=19.6 N and hk=0.40×1960=784 N ∴ x=(19.6±√((19.6)2+2(19.6)(784)))/1960 =0.10 m or -0.080 m Since, x must be positive (a compression) we accept the positive solution and reject the negative solution. Hence, x=0.10 m
Let m be the mass of the block, h the height from which it is dropped, and x the compression o the spring. Since, energy is conserved, so Final gravitational potential energy = final spring potential energy or mg(h+x)=1/2 kx2 or mg(h+x)+1/2 kx2=0 or kx2-2mg(h+x)=0 kx2-2mgx-2mgh=0 This is a quadratic equation for x. Its solution is x=(mg± √((mg)2+2mghk))/k Now, mg=2×9.8=19.6 N and hk=0.40×1960=784 N ∴ x=(19.6±√((19.6)2+2(19.6)(784)))/1960 =0.10 m or -0.080 m Since, x must be positive (a compression) we accept the positive solution and reject the negative solution. Hence, x=0.10 m
Q6. A ball is dropped from height 20 m. If coefficient of restitution is 0.9, what will be the height attained after first bounce?
Solution
If a body falls from height h, then from equation of motion we know that it will hit the ground with a velocity say u=√2gh which is also the velocity of approach here. Now, if after collision it gains a height h1 then again by equation of motion v=√2gh, which is also the velocity of separation .so, by definition of e, e=√((2gh1)/2gh) or h1=e2 h Given ,h=20 m, e=0.9 ∴ height attained after first bounce h1=(0.9)2×20 =0.9×0.9×20 =16.2
If a body falls from height h, then from equation of motion we know that it will hit the ground with a velocity say u=√2gh which is also the velocity of approach here. Now, if after collision it gains a height h1 then again by equation of motion v=√2gh, which is also the velocity of separation .so, by definition of e, e=√((2gh1)/2gh) or h1=e2 h Given ,h=20 m, e=0.9 ∴ height attained after first bounce h1=(0.9)2×20 =0.9×0.9×20 =16.2
Q7. A car is moving with a speed of 100 kmh-1. If the mass of the car is 950 kg, then its kinetic energy is?
Solution
Kinetic energy, =1/2×950×(100×5/18)2 J =0.3665×106 J = 0.367 MJ
Kinetic energy, =1/2×950×(100×5/18)2 J =0.3665×106 J = 0.367 MJ
Q8. A body of mass 2 kg moving with a velocity of 3 m/sec collides head on with a body of mass 1 kg moving in opposite direction with a velocity of 4 m/sec. After collision, two bodies stick together and move with a common velocity which in m/sec is equal to ?
Solution
m1 v1-m2 v2=(m1+m2 )v ⇒2×3-1×4=(2+1)v ⇒v=2/3 m/s
m1 v1-m2 v2=(m1+m2 )v ⇒2×3-1×4=(2+1)v ⇒v=2/3 m/s
Q9. A particle is moving under the influence of a force given by F=kx where k is a constant and x is the distance moved. The energy (in joules) gained by the particle in moving from x=0 to x=3 is?
Solution
U= 1/2 K(x22-x12 )⇒U=1/2 K(32-0)⇒U=4.5 K
U= 1/2 K(x22-x12 )⇒U=1/2 K(32-0)⇒U=4.5 K
Q10. If a man speeds up by 1ms-1), his KE increase by 44%. His original speed in ms-1 is?:
Solution
E1=1/2 mv2 E2=1/2 m(v+1)2 ((E2-E1))/E1 =(1/2 m[(v+1)2-v2])/(1/2 mv2 )=44/100 On solving, we get v=5ms-1)
E1=1/2 mv2 E2=1/2 m(v+1)2 ((E2-E1))/E1 =(1/2 m[(v+1)2-v2])/(1/2 mv2 )=44/100 On solving, we get v=5ms-1)
