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On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle 0 to its initial direction and has a speed v/3. The second block's speed after the collision
- a)3/√2v
- b)√3/2v
- c)2√2/3v
- d)3/4v
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
On a frictionless surface, a block of mass M moving at speed v collid...
The situation is shown in the figure.
Let V be speed of second block after the collision.
As the collision is elastic, so kinetic energy is conserved.
According to conservation of kinetic energy,
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Community Answer
On a frictionless surface, a block of mass M moving at speed v collid...
Collision Analysis:
In an elastic collision, both momentum and kinetic energy are conserved. Let's analyze the collision step by step.
Initial Conditions:
- Block 1 has mass M and initial velocity v.
- Block 2 has mass M and is initially at rest.
Step 1: Conservation of Momentum
Since the collision is elastic, the total momentum before and after the collision must be the same.
Initial momentum (before collision):
P_initial = M * v
Final momentum (after collision):
P_final = P1_final + P2_final
Since Block 2 is initially at rest, its final momentum is zero:
P2_final = 0
Therefore, P_final = P1_final.
Step 2: Collision Analysis
Let's consider the collision from a center-of-mass frame of reference.
In this frame, the initial velocity of the center of mass is zero, and both blocks have equal and opposite velocities relative to the center of mass.
After the collision, the first block moves at an angle θ to its initial direction with a speed of v/3.
Using trigonometry, we can express the final velocities of both blocks in terms of the angle θ:
Block 1 final velocity (V1_final):
V1_final = (v/3) * cos(θ) (horizontal component)
Block 2 final velocity (V2_final):
V2_final = (v/3) * sin(θ) (vertical component)
Step 3: Conservation of Kinetic Energy
Since the collision is elastic, the total kinetic energy before and after the collision must be the same.
Initial kinetic energy (before collision):
KE_initial = (1/2) * M * v^2
Final kinetic energy (after collision):
KE_final = (1/2) * M * V1_final^2 + (1/2) * M * V2_final^2
Substituting the expressions for V1_final and V2_final:
KE_final = (1/2) * M * [(v/3) * cos(θ)]^2 + (1/2) * M * [(v/3) * sin(θ)]^2
Simplifying the equation:
KE_final = (1/2) * M * (v^2/9) * [cos^2(θ) + sin^2(θ)]
Using the trigonometric identity cos^2(θ) + sin^2(θ) = 1:
KE_final = (1/2) * M * (v^2/9)
Since KE_initial = KE_final, we can equate the expressions:
(1/2) * M * v^2 = (1/2) * M * (v^2/9)
Simplifying the equation:
1 = 1/9
Therefore, the only possible value for the angle θ is 45°.
Step 4: Finding the Speed of Block 2
Now that we know the angle θ, we can find the final velocity of Block 2.
Using the expression for V2_final:
V2_final = (v/3) * sin(45°)
Simplifying the equation:
V2_final = (v/3) * (√2/2) = (√2/6) * v
Therefore, the correct answer is option C: 2√2/3v
In an elastic collision, both momentum and kinetic energy are conserved. Let's analyze the collision step by step.
Initial Conditions:
- Block 1 has mass M and initial velocity v.
- Block 2 has mass M and is initially at rest.
Step 1: Conservation of Momentum
Since the collision is elastic, the total momentum before and after the collision must be the same.
Initial momentum (before collision):
P_initial = M * v
Final momentum (after collision):
P_final = P1_final + P2_final
Since Block 2 is initially at rest, its final momentum is zero:
P2_final = 0
Therefore, P_final = P1_final.
Step 2: Collision Analysis
Let's consider the collision from a center-of-mass frame of reference.
In this frame, the initial velocity of the center of mass is zero, and both blocks have equal and opposite velocities relative to the center of mass.
After the collision, the first block moves at an angle θ to its initial direction with a speed of v/3.
Using trigonometry, we can express the final velocities of both blocks in terms of the angle θ:
Block 1 final velocity (V1_final):
V1_final = (v/3) * cos(θ) (horizontal component)
Block 2 final velocity (V2_final):
V2_final = (v/3) * sin(θ) (vertical component)
Step 3: Conservation of Kinetic Energy
Since the collision is elastic, the total kinetic energy before and after the collision must be the same.
Initial kinetic energy (before collision):
KE_initial = (1/2) * M * v^2
Final kinetic energy (after collision):
KE_final = (1/2) * M * V1_final^2 + (1/2) * M * V2_final^2
Substituting the expressions for V1_final and V2_final:
KE_final = (1/2) * M * [(v/3) * cos(θ)]^2 + (1/2) * M * [(v/3) * sin(θ)]^2
Simplifying the equation:
KE_final = (1/2) * M * (v^2/9) * [cos^2(θ) + sin^2(θ)]
Using the trigonometric identity cos^2(θ) + sin^2(θ) = 1:
KE_final = (1/2) * M * (v^2/9)
Since KE_initial = KE_final, we can equate the expressions:
(1/2) * M * v^2 = (1/2) * M * (v^2/9)
Simplifying the equation:
1 = 1/9
Therefore, the only possible value for the angle θ is 45°.
Step 4: Finding the Speed of Block 2
Now that we know the angle θ, we can find the final velocity of Block 2.
Using the expression for V2_final:
V2_final = (v/3) * sin(45°)
Simplifying the equation:
V2_final = (v/3) * (√2/2) = (√2/6) * v
Therefore, the correct answer is option C: 2√2/3v
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On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle 0 to its initial direction and has a speed v/3. The second block's speed after the collisiona)3/√2vb)√3/2vc)2√2/3vd)3/4vCorrect answer is option 'C'. Can you explain this answer?
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