Answer:
The rise in height of combined block/bullet from its original position is 0.45m
Explanation:
Given;
mass of bullet, m₁ = 12 g = 0.012 kg
mass of block of wood, m₂ = 1 kg
initial speed of bullet, u₁ = 250 m/s.
initial speed of block of wood, u₂ = 0
From the principle of conservation of linear momentum, calculate the final speed of the combined block/bullet system.
m₁u₁ + m₂u₂ = v(m₁+m₂)
where;
v is the final speed of the combined block/bullet system.
0.012 x 250 + 0 = v (0.012 + 1)
3 = v (1.012)
v = 3/1.012
v = 2.96 m/s
From the principle of conservation of energy, calculate the rise in height of the block/bullet combined from its original position.
¹/₂mv² = mgh
¹/₂v² = gh
¹/₂ (2.96)² = (9.8)h
4.3808 = 9.8h
h = 4.3808/9.8
h = 0.45 m
Therefore, the rise in height of combined block/bullet from its original position is 0.45m
Answer:
0.45 m
Explanation:
From the law of conservation of momentum,
mv = (m + M)V where m = mass of bullet = 12 g = 0.012 kg, v = initial speed of bullet = 250 m/s, M = mass of block = 1 kg, V = final velocity of bullet and block.
V = mv/(m + M) = 0.012 kg 250 m/s/(0.012 + 1) kg = 3/1.012 = 2.96 m/s
From the law of conservation of energy,
ΔU + ΔK = 0 where U = potential energy of mass and block and K = kinetic energy of mass and block
ΔU = -ΔK
(m + M)gh - 0 = -[0 - 1/2(m + M)V²]
(m + M)gh - 0 = -[0 - 1/2(m + M)V²]
(m + M)gh = 1/2(m + M)V²
gh = 1/2V²
h = V²/2g where h is the height moved by the combined bullet and block
h = (2.96 m/s)²/(2 × 9.8 m/s²)
h = 0.45 m
