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.?

Question

Answer

See ayushi, the angle always remains 90 degree. that means after collision the total angle made by the bodies will be 90 degree. since the body A is making an angle of 37 degree, body B will.make an angle of 53 degree. now the remaining question for the speeds, you can apply simple conservation of momentum along with cosine angles.mu= m1u1cos37 + m2u2cos53.cos37= 4/5. cos53=3/5

Answer

Problem Statement

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 the minimum possible speed of B for different possible collisions?

Solution

As the collision is oblique, we need to resolve the velocities of both A and B before and after the collision in the direction of collision and perpendicular to it.

Initial velocities:

Velocity of ball A in the direction of collision = 4cos(theta)

Velocity of ball A perpendicular to the direction of collision = 4sin(theta)

Velocity of ball B in the direction of collision = 0

Velocity of ball B perpendicular to the direction of collision = 0

Final velocities:

Velocity of ball A in the direction of collision = Vcos(37)

Velocity of ball A perpendicular to the direction of collision = Vsin(37)

Velocity of ball B in the direction of collision = Vb

Velocity of ball B perpendicular to the direction of collision = 0

Conservation of Momentum:

As the collision is inelastic, we can use the law of conservation of momentum to find the final velocity of ball A:

Initial momentum = Final momentum

10*4cos(theta) = 10Vcos(37) + 8Vb

Conservation of Energy:

As the collision is inelastic, we can use the law of conservation of energy to find the final velocity of ball A:

Initial Kinetic Energy = Final Kinetic Energy

0.5*10*(4)^2 = 0.5*10*(Vcos(37))^2 + 0.5*8*(Vb)^2

Calculation:

Using the above two equations, we can solve for V and Vb:

V = 2.174 m/s

Vb = 1.304 m/s

Conclusion:

The minimum possible speed of ball B for different possible collisions is 1.304 m/s.

Answer

84 degree

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