A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood was hanging from a 2 m long piece of (massless) string. After the collision the block/bullet combined object swings upward on the string. Find the height the block/bullet combined object rises.

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

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

chemical properties of thing

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s