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Question
A sphere of mass 10 kg is moving on a horizontal plane with a velocity (10^i+^j) m/s where it collides with a vertical wall which is parallel to ^j. If the coefficient of restitution is 0.5, find the impulse (in SI units) acted on the sphere.
A sphere of mass 10 kg is moving on a horizontal plane with a velocity (10^i+^j) m/s where it collides with a vertical wall which is parallel to ^j. If the coefficient of restitution is 0.5, find the impulse (in SI units) acted on the sphere.
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Solution
The correct option is C −150 ^i
Given, velocity of the sphere →v=10^i+^j.
The situation is as shown in the below figure.
Momentum of the individual particle will be conserved perpendicular to LOI. Here along ^j (along the vertical wall) momentum of individual particle will be conserved.
As, e=Velocity of seperationVelocity of approach along LOI
⇒ 0.5=0−vx10−0
⇒ vx=−5^i where, vx is the speed of sphere after collision, along the x−axis or along the LOI.
Hence, velocity of sphere after collision is
→v=−5^i+^j
So, impulse on sphere is given by
→J=△P=m→v−m→u
⇒ →J=m[(−5^i+^j)−(10^i+^j)]=10×[−15 ^i]
⇒ →J=−150 ^i
Given, velocity of the sphere →v=10^i+^j.
The situation is as shown in the below figure.
Momentum of the individual particle will be conserved perpendicular to LOI. Here along ^j (along the vertical wall) momentum of individual particle will be conserved.
As, e=Velocity of seperationVelocity of approach along LOI
⇒ 0.5=0−vx10−0
⇒ vx=−5^i where, vx is the speed of sphere after collision, along the x−axis or along the LOI.
Hence, velocity of sphere after collision is
→v=−5^i+^j
So, impulse on sphere is given by
→J=△P=m→v−m→u
⇒ →J=m[(−5^i+^j)−(10^i+^j)]=10×[−15 ^i]
⇒ →J=−150 ^i
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