Collision, Spring, Energy Balance

Q. Collision, Spring, Energy Balance

Question. 

Body A, of mass 2kg, has a light spring attached to it as shown in the adjoining figure.

Body B has a mass of 5kg and is initially suspended by a string of length 0.1m. Body A moves with a velocity of 3.5m/s over a frictionless plane directly towards body B as shown in Fig. 1.1. The two bodies undergo an elastic collision, during which the spring on body A is compressed as shown in Fig. 1.2 and then the spring extended. The two bodies separate after collision as shown in Fig. 1.3

i)During the maximum compression, the initial kinetic energy of A is more than the total kinetic energy of A and B. Explain why this is possible.

ii)Find the velocities VA and VB after the collision.

iii)Calculate the tension of the string when B swings through an angle of 30 degrees and reaches point C (Take g=10m/s^2).

Answer .

i)

During the maximum compression, kinetic energy of A is converted into the kinetic energy of B plus potential energy stored in the spring. Thus, kinetic energy of A is zero and of its original kinetic energy only a part is converted into the kinetic energy of B and the balance is stored as the potential energy in the spring. This is the reason why original kinetic energy of A is more than the sum of the kinetic energy of A at maximum compression of the spring (which is zero) and the kinetic energy of B (which is only a part of the kinetic en energy of A).

(ii)
By the law of conservation of linear momentum,

2 * 3.5 = 2 * (- VA) + 5 * VB

=> VA = 2.5VB - 3.5 ... ( 1 )

As A moves on frictionless plane and collision is elastic, kinetic energy is also conserved.

=> (1/2) * 2 * (3.5)^2 = (1/2) * 2 * VA^2 + (1/2) * 5 * VB^2

=> 2.5 VB^2 = (3.5)^2 - VA^2

Plugging the value of VA from ( 1 ),

=> 2.5VB^2 = 12.25 - (2.5VB - 3.5)^2

=> 2.5VB^2 = 17.5VB - 6.25VB^2

=> 8.75VB = 17.5

=> VB = 2 m/s.

and VA = 2.5VB - 3.5 = 2.5 * 2 - 3.5 = 1.5 m/s.

(iii)

When B goes to point C, it rises through height, h

= (0.1) (1 - cos30°) and gains PE

= mgh

= 5*10*(0.1)(1 - cos30°) J

= 0.67 J

KE at C = KE at B - 0.67

=> (1/2) * 5 * v^2 = (1/2) * 5 * (2)^2 - 0.67

=> v^2 = 2.68 J

If T = tension in the string, balancing forces radially,

T - mgcos30° = mv^2/r = 5 *2.68/(0.1)

=> Tension, T

= 5*10*cos30° + 5*26.8 N

= 177.3 N.