A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s.

Question

Answer

Answer

Velocity of the first block before collision (u1) = 10 m/s
Mass of the first block (m1) = 4 kg
Velocity of the second block before collision (u2) = 0 m/s (at rest)
Mass of the second block (m2) = 5 kg

The collision between the two blocks is elastic, which means both momentum and kinetic energy are conserved.

1. Conservation of momentum:
Before collision: m1u1 + m2u2 = m1v1 + m2v2
After collision, the two blocks move in different directions. Let's take the positive direction as the direction in which the first block moves.

m1u1 - m2u2 = m1v1 - m2v2 (Equation 1)

2. Conservation of kinetic energy:
Before collision: (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Since the collision is elastic, the kinetic energy is conserved.

(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2 (Equation 2)

3. Solving for the velocities:
Using equation 1, we can solve for v1:

m1u1 - m2u2 = m1v1 - m2v2
4(10) - 5(0) = 4v1 - 5v2
40 = 4v1 - 5v2 (Equation 3)

Using equation 2, we can solve for v2:

(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
2(100) = 4v1^2 + 5v2^2
200 = 4v1^2 + 5v2^2 (Equation 4)

4. Solving the equations:
Multiplying equation 3 by 4 and equation 4 by 5, we can eliminate the v2 term:

160 = 16v1 - 20v2
1000 = 20v1^2 + 25v2^2

Multiplying equation 3 by 5 and equation 4 by 4, we can eliminate the v1 term:

200 = 20v1 - 25v2
800 = 16v1^2 + 20v2^2

Adding the two resulting equations:

160 + 200 = 16v1 - 20v2 + 20v1 - 25v2
360 = 36v1 - 45v2

Rearranging the equation:

36v1 - 45v2 = 360

Using the angle of deflection, we can express v1 and v2 in terms of trigonometric functions:

v1 = v cos(30)
v2 = v sin(60)

Substituting these values into the equation:

36v cos(30) - 45v sin(60) = 360

Simplifying:

36(√3/2)v - 45(√3/2)v

Similar Class 12 Doubts

A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.?

Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain

A body of mass 5 kg make an elastic collision with another body at rest and continue to move in the originna direction after collision with a velocity equal is 1/10th of the original velocity then mass of the second body is?

A body of mass 5 kg an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal is 1/10th of the original velocity .then mass of the second body is?

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams