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Usimov [2.4K]
3 years ago
6

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.
Physics
2 answers:
LuckyWell [14K]3 years ago
7 0

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

topjm [15]3 years ago
5 0

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

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Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

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F = -kQ3(Q1/(r1)² + Q2/(r2)²)

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F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

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3 years ago
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