Motion, Energy balance, momentum question

[Raghav Gupta]

Homework Statement

A block of mass m= 10kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05.
When hit by a bullet of mass 50 g moving with speed v, that gets embedded in it, the block moves and comes to stop after moving a distance of 2m on the table.
If a freely falling object were to acquire speed v/10 after being dropped from height H, then neglecting energy losses and taking g= 10 m/s², the value of H is close to:
A. 0.2 km
B. 0.3 km
C. 0.4 km
D 0.5. km

Homework Equations

Momentum conservation, energy conservation, Newton laws

The Attempt at a Solution

$$M_{bullet} v = ( m_{block} + M_{bullet} )v_{block} $$ ( momentum conservation)

$$ 1/2 (m_{block} + M_{bullet}) v_{block}^2 = μ (m_{block} + M_{bullet}) g s$$ energy conservation
$$ v_{block}^2 = 2μgs = 2 * 0.05 * 10 * 2 = 2 $$
$$ v_{block} = √2 m/s $$
$$ v = ( m_{block} + M_{bullet} )v_{block}/M_{bullet} ≈ 10000 * √2/50 = 200√2 m/s $$
$$ v/10 = 20√2 $$

Now, ## mgH = 1/2 m (v/10)^2 ##
## ⇒ H = (20√2)^2/2g = 800/20 = 40 m = 0.04 km ##
What wrong I have done here?
It is not matching with options.

 

[ehild]

What options are there?
I think you rounded off too early.

 

[Raghav Gupta]

What options are there?
I think you rounded off too early.

0.4 km
0.3 km
0.2 km
0.5 km
But why so error in rounding?
By a factor of 10?

 

[ehild]

It is not rounding error. Your result is correct, the problem maker has error.