Question

1. A block of mass 0.4kg resting on the top of an inclined plane of height 20m starts to slide down on the surface of the incline to its foot, and then continues its slide horizontally. At a distance of 5m from the foot of the incline there is another block of the same mass resting on the horizontal surface to undergo an elastic collision. Next to the second block, there is a light spring of constant k = 4000N/m fixed freely against a wall. The spring is supposed to make a head-on collision with the second block. See the arrangements as in 1. Assuming all surfaces being frictionless, (a) calculate the kinetic energy of the firs block just at the foot of the incline; (b) calculate the kinetic and gravitational potential energies of the first block halfway down the incline; (c) calculate the speeds of the two blocks just after their collision; (d) compute the maximum compression of the spring resulted from its collision with the second block; (e) determine the maximum work done by the spring on the second block.​

Answer 1

The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

To find the answer, we need to know about the concept of collision and kinetic energy.

How to find the kinetic energy of the first block just at the foot of the incline?

• Given that, the block of mass 0.4kg resting on the top of an inclined plane of height 20m.
• Thus, at the top of the incline it has a potential energy, and the kinetic energy will be equal to zero, or we can say that the total energy of the system is equal to the potential energy at topmost point.

                 $TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J$

• We have to find the kinetic energy of the first block just at the foot of the incline, and at the bottom point the PE=0, or we can say that the total energy or the potential energy is converted into kinetic energy.

                   $TE=KE=78.4J$

What is the kinetic and gravitational potential energies of the first block halfway down the incline?

• At the halfway, the PE will be,

                          $U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J$

• As we know that, the energy is conserved at each point of the motion.

                      $TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J$

How to find the speeds of the two blocks just after their collision?

• We have the KE at bottom point as, 78.4J. Thus, the velocity of first block at the bottom before collision will be,

                            $KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s$

• This is the velocity of the block 1 of mass m1 before collision, we can say, u1.
• As we know that, the 2 nd block of mass m2 is at rest, thus, u2=0.
• Given that, the collision is elastic. Thus, both the KE and the momentum will be conserved.

               $\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

                 $m_1u_1+m_2u_2=m_1v_1+m_2v_2$

• We have,

                            $m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?$

• Substituting this in both the equations, we get,

                       $\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16$  from resolving KE equation.

                     

                        $4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2$ From resolving momentum conservation.

• solving both, we get,

                            $v_2=4m/s\\v_1=0$

Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

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