I came across this question:

A pellet with velocity 200 ms⁻¹ and mass 5.0 g is fired vertically upwards into a stationary block of mass 95.0 g. The pellet remains in the block. The impact causes the block to move vertically upwards.

What is the maximum vertical displacement of the block?

I initially used conservation of energy to (incorrectly) answer this question:

KE of pellet = GPE of block

½(5x10⁻³)(200²) = (95x10⁻³+5x10⁻³)(9.81)Δh

⇒ Δh = 101.9 m

I thought the above value looked too large, so I re-attempted the question using conservation of momentum:

momentum of pellet = momentum of block

(5x10⁻³)(200) = (95x10⁻³+5x10⁻³)v

⇒ v = 10 ms⁻¹

v² = u²+2as

0 = 10²+2(-9.81)s

⇒ s = 5.1 m

Which is the right answer.

What did I do wrong in the first approach?

I'm guessing I made an assumption which means I can't use conservation of energy to answer the question. I mean yes some energy is lost as heat and sound and whatnot in the collision but I didn't think it'd make such a huge difference in the result.

A pellet with velocity 200 ms⁻¹ and mass 5.0 g is fired vertically upwards into a stationary block of mass 95.0 g. The pellet remains in the block. The impact causes the block to move vertically upwards.

What is the maximum vertical displacement of the block?

I initially used conservation of energy to (incorrectly) answer this question:

KE of pellet = GPE of block

½(5x10⁻³)(200²) = (95x10⁻³+5x10⁻³)(9.81)Δh

⇒ Δh = 101.9 m

I thought the above value looked too large, so I re-attempted the question using conservation of momentum:

momentum of pellet = momentum of block

(5x10⁻³)(200) = (95x10⁻³+5x10⁻³)v

⇒ v = 10 ms⁻¹

v² = u²+2as

0 = 10²+2(-9.81)s

⇒ s = 5.1 m

Which is the right answer.

What did I do wrong in the first approach?

I'm guessing I made an assumption which means I can't use conservation of energy to answer the question. I mean yes some energy is lost as heat and sound and whatnot in the collision but I didn't think it'd make such a huge difference in the result.

Original post by notacountry

I came across this question:

A pellet with velocity 200 ms⁻¹ and mass 5.0 g is fired vertically upwards into a stationary block of mass 95.0 g. The pellet remains in the block. The impact causes the block to move vertically upwards.

What is the maximum vertical displacement of the block?

I initially used conservation of energy to (incorrectly) answer this question:

KE of pellet = GPE of block

½(5x10⁻³)(200²) = (95x10⁻³+5x10⁻³)(9.81)Δh

⇒ Δh = 101.9 m

I thought the above value looked too large, so I re-attempted the question using conservation of momentum:

momentum of pellet = momentum of block

(5x10⁻³)(200) = (95x10⁻³+5x10⁻³)v

⇒ v = 10 ms⁻¹

v² = u²+2as

0 = 10²+2(-9.81)s

⇒ s = 5.1 m

Which is the right answer.

What did I do wrong in the first approach?

I'm guessing I made an assumption which means I can't use conservation of energy to answer the question. I mean yes some energy is lost as heat and sound and whatnot in the collision but I didn't think it'd make such a huge difference in the result.

A pellet with velocity 200 ms⁻¹ and mass 5.0 g is fired vertically upwards into a stationary block of mass 95.0 g. The pellet remains in the block. The impact causes the block to move vertically upwards.

What is the maximum vertical displacement of the block?

I initially used conservation of energy to (incorrectly) answer this question:

KE of pellet = GPE of block

½(5x10⁻³)(200²) = (95x10⁻³+5x10⁻³)(9.81)Δh

⇒ Δh = 101.9 m

I thought the above value looked too large, so I re-attempted the question using conservation of momentum:

momentum of pellet = momentum of block

(5x10⁻³)(200) = (95x10⁻³+5x10⁻³)v

⇒ v = 10 ms⁻¹

v² = u²+2as

0 = 10²+2(-9.81)s

⇒ s = 5.1 m

Which is the right answer.

What did I do wrong in the first approach?

I'm guessing I made an assumption which means I can't use conservation of energy to answer the question. I mean yes some energy is lost as heat and sound and whatnot in the collision but I didn't think it'd make such a huge difference in the result.

Its an inelastic collision so the maximum amount of KE is lost

https://en.wikipedia.org/wiki/Inelastic_collision#:~:text=A%20perfectly%20inelastic%20collision%20occurs,bonding%20the%20two%20bodies%20together.

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