Question

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 12.0 g is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 4.80 N. What was the initial speed v0 of the bullet?

Answer 1

Answer:

The initial velocity of the bullet is

Vo =281.52 m/s

Explanation:

Given that,

Mass of block, M = 0.8kg

Then, weight of the block is

W = Mg = 0.8 ×9.81= 7.848N

Length of cord use to suspend mass is

l = 1.6m

Mass of bullet fired is

m=12g = 12/1000 =0.012kg

The initial velocity of the bullet is Vo

But the block is at rest before the bullet was fired, then, it's initial velocity is Uo =0m/s

When the bullet hits the block it moves to a height of h=0.8m

Tension in cord when the body rise to that height is T = 4.8N

So we want to find the initial velocity is Vo.

Using conservation of momentum

Momentum before collision is equal to the momentum after collision

Momentum is given as p=mv

The momentum before collision of the bullet and the block is

P(initial) = mVo + MUo

P(initial) = 0.012×Vo + 0.8 × Uo

P(initial) = 0.012Vo

After collision, both the block and the mass are moving together, i.e. the inelastic collision

The final momentum after collision is given as

P(final) = (m+M)V1

P(final) =(0.012+0.8)V1

P(final) = 0.812V1

Now, we need to know V1.

To know the angle the mass suspended after being fired.

Check attachment for better view

Then, using trigonometry

Cosθ=adj/hyp

Cosθ=0.8/1.6

Cosθ=0.5

θ=arccos(0.5)

θ=60°

Applying Newton second law of motion at the point 2 along radial direction

ΣF = ma

T — W2•Cosθ = (m+M)a

Where a is radial acceleration and it is given as a=v²/r. r=l

T — (m+M)g•Cosθ = (m+M)a

4.8—(0.012+0.8)×9.81•Cos60=(0.012+0.8)V2²/l

4.8 — 3.983=0.812V2²/1.6

0.876 = 0.812V2²/1.6

0.81714 = 0.5075V2²

V2² = 0.81714/0.5075

V2²= 1.61

V2 = √1.61

V2 = 1.269 m/s

V2 ≈ 1.27m/s

This is the velocity of the object and the bullet at point 2

But, we need V1

Let analyse point 1

Using conservation of energy

i.e total energy at point 1 to energy at point 2.

K1 + U1 = K2 + U2

Potential energy is given as mgh

Kinetic energy is given as ½mv²

Note that U1 is zero because change in height is zero

Then, we have

K1 = K2 + U2

½(m+M)V1² =½(m+M)V2²+(m+M)•g•h

½(0.012+0.8)V1² = ½(0.012+0.8)1.27² + (0.012+0.8)•9.81•0.8

0.406V1² = 0.6548 + 6.373

0.406V1² = 7.0274

V1² = 7.0274/0.406

V1² = 17.309

V1 = √17.309

V1 = 4.16m/s

Then, applying it to the conservation of momentum

P(initial) = P(final)

0.012Vo = 0.812V1

0.012Vo = 0.812 ×4.16

0.012Vo = 3.378

Vo = 3.378/0.012

Vo = 281.52 m/s

Then, the initial velocity of the bullet is 281.52m/s