Kinetic/potential energy problem?

Question

Could you please show me how to do this? i have the answer in my textbook but i have no idea how to get it.

Jane threw a dark of mass 50g into a block of soft wood. The block of mass 1kg was suspended freely by a length of fishing line before being hit by the dart. Andrew observed that the block with the dark in it swung to one side and rose a vertical distance of 40cm. Calculate the speed of the dart when it hit the block.

The answer is 32.3. Any help is appreciated :]

Answer 1

Let

u = speed of dart when it hit the block

v = speed of dart + block after the dart hits the block

m1 = mass of dart

m2 = mass of block

h = height to which dart + bock rises

Just after the dart gets embedded in the block,

KE of block = dart = 1/2 * (m1 + m2)v^2

When the system rises to maximum height h, then KE becomes zero and PE rises by (m1 + m2)gh

By conservation of energy,

1/2 * (m1 + m2)v^2 = (m1 + m2)gh

Divide by m1+m2 on both sides

1/2 * v^2 = gh

Or v = sqrt(2gh)-------------(1)

Just before the dart hits the block,

momentum of the system = momentum of dart + momentum of block

= m1u + 0  (because the block is at rest)

= m1u

Just after the dart hits the block,

momentum of the system = (m1+m2)v  = (m1+m2)*sqrt(2gh) [using (1)]

By conservation of momentum

m1u = (m1+m2)*sqrt(2gh)

Or u = (m1+m2)*sqrt(2gh)/m1

m1 = 50 g = 0.05 kg

m2 = 1 kg

h = 40 cm = 0.4 m

Therefore

u = (0.05 + 1)*sqrt(2*9.8*0.4)/0.05 m/s

= 1.05 * sqrt(7.84)/0.05 m/s

= 1.05 * 2.8/0.05 m/s

= 58.8 m/s

Ans: 58.8 m/s

Are you sure that the answer is 32.3? And what is the unit?

Answer 2

KEi=PEf

1/2mv^2=mgh

(1/2)(1050g)(v^2)=(1050g)(980cm/s/s)(4...

v=280cm/s

p(block and dart)=p(dart) + p(block)

(mdart+mblock)v=mdart(v) + 0

(1050g)(280cm/s)=50g(v)

5880 cm/s=v

v(dart)=58.8 m/s

=192.92 ft/s