A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9·8 m/s²)
- (a)980 m s⁻¹ and 980 J
- (b)√980 m s⁻¹ and √980 J
- (c)980 m s⁻¹ and √980 J
- (d)√980 m s⁻¹ and 980 J
Correct — D, sqrt(980) m/s and 980 J. The body is dropped from rest, so its total energy at the top is purely potential: E = mgh = 2 x 9.8 x 50 = 980 J. As it falls this energy converts entirely into kinetic energy at the ground: (1/2)mv^2 = mgh, giving v = sqrt(2gh) = sqrt(2 x 9.8 x 50) = sqrt(980) m/s (about 31.3 m/s). So the impact speed is sqrt(980) m/s and the total energy is 980 J.
- (a)980 m s(-1) and 980 J — The total energy 980 J is right, but the speed is wrong. v = sqrt(2gh) = sqrt(980) is about 31.3 m/s, not 980 m/s - the number 980 is the value of 2gh (i.e. v-squared), not v.
- (b)sqrt(980) m s(-1) and sqrt(980) J — The speed sqrt(980) m/s is right, but the energy is wrong. The total energy is mgh = 980 J, not sqrt(980) J.
- (c)980 m s(-1) and sqrt(980) J — Both quantities are wrong - the speed is sqrt(980) m/s (not 980) and the total energy is 980 J (not sqrt(980)).
For a body dropped from rest under gravity, mechanical energy is conserved. Its total energy at the release point equals its potential energy, mgh. On reaching the ground all of it becomes kinetic energy, so (1/2)mv^2 = mgh, which gives the impact speed v = sqrt(2gh) - independent of the mass.
Compute the two quantities separately. Total energy when released (at rest) is just PE = mgh = 2 x 9.8 x 50 = 980 J. Impact speed comes from equating KE to PE: v = sqrt(2gh) = sqrt(980) m/s. The trap is to report 980 (which is v-squared) as the speed - keep the square root.
- For a body dropped from rest, v = sqrt(2gh) (independent of mass).
- Total mechanical energy at the top equals the potential energy, mgh.
- Energy is conserved: PE at the top fully becomes KE at the bottom (no air resistance).
- Here E = 2 x 9.8 x 50 = 980 J and v = sqrt(2 x 9.8 x 50) = sqrt(980), about 31.3 m/s.
- Confusing v-squared (= 2gh = 980) with v (= sqrt(980)) - 980 is the speed squared, not the speed.
- Thinking heavier bodies fall faster - v = sqrt(2gh) does not depend on mass.
Free-fall numericals ask for the impact speed (sqrt(2gh)) and/or the energy (mgh) - plug in the numbers and keep the square root.
No directly related past PYQ was found.
- practice — not a real PYQ
A body is dropped from a height h. Its speed on reaching the ground is
- (a)sqrt(gh)
- (b)sqrt(2gh)
- (c)2gh
- (d)gh
Answer(b) sqrt(2gh) - from energy conservation, (1/2)mv^2 = mgh.
- practice — not a real PYQ
A 1 kg ball is at a height of 10 m (g = 10 m/s^2). Its potential energy is
- (a)10 J
- (b)100 J
- (c)1 J
- (d)1000 J
Answer(b) 100 J - PE = mgh = 1 x 10 x 10.