Two bodies of masses M and m(M>m) are allowed to fall from the same height. If
1 answer:
- Two bodies of masses M and m (M>m) are allowed to fall from the same height . If the air resistance for each be the same then both the bodies will reach the earth simultaneously .
- For same air resistance, acceleration of both the bodies will be same .
[A] If both assertion and reason are true and reason is the correct explanation of assertion .
[B] If both assertion and reason are true but reason is not the correct explanation of assertion .
[C] If assertion is true but reason is false
[D] If both assertion and reason are false
✍️ The force acting on the body of mass M are [its weight] ‘<u>Mg</u>’
- Mg are acting vertically downwards
- And air resistance F acting vertically upward .
Hence,
⚡ Now, M > m therefore, the body with larger mass will have greater acceleration and it will reach the ground first .
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Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = ( × mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s