Find the maximum height, in cm, the large block reaches.

Question

A small block of mass 210 grams starts from rest at a height of 260 cm on a frictionless track, as shown. It is released and slides down the track, colliding with a large block of mass 566 grams. The small block rebounds and reaches a maximum height of 45.0 cm, while the large block slides to the opposite end of the track. Find the maximum height, in cm, the large block reaches. 

Answer 1

Given

Small block:

Small block:

1. m₁=210 g = 0.210 kg
2. Initial height h_1i = 260 cm= 2.60 m
3. Rebound height h_1f = 45.0 cm = 0.45 m

Large block:

1. m₂=566 g=0.566 kg
2. Initial height = 0
3. Final height = ?

Track is frictionless → mechanical energy conserved during motion.

Collision happens at the bottom.

Step 1: Speed of small block before collision

Using energy conservation

m₁gh_1i = 1/2 m₁ v_1i²

mass cancels from both sides of equation

then we get v_1i​= Square root of ( 2gh_1i)

​​ v1i​ = square root of ( 2x9.8x2.6) = 7.14 m/s

Step 2: Speed of small block after collision

v_1f​ ² = 2gh_1f​​ = 2x9.8x0.45 hence v_1f​ = square root of ( 2x9.8x 0.45) = 2.97 m/s

since it rebounds , direction is opposite so v_1f​ = - 2.97 m/s

Step 3: Use conservation of momentum

Before collision:

p_i = m₁v_1i

​After collision:

p_f = m₁v_1f + m₂v_2f

​So, p_i = p_f

m₁v_1i = m₁v_1f + m₂v_2f

(0.210)(7.14)=(0.210)(−2.97)+(0.566)v_2f​

2.123=0.566v_2f​

v_2f​ = 3.75 m/s

Step 4: Convert this to maximum height

All kinetic energy converts to gravitational potential:

0.5 m₂v_2f² = m₂gh₂ h₂ = v_2f² / 2g

h₂ = (3.75)² / 2x9.8 = 0.717 m = 71.7cm

Ans: The large block reaches a maximum height of approximately 72 cm.