Solving the Basic Mechanics Bullet Block Collision Problem

[Arm]

Homework Statement

A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block which is
resting on a horizontal table. The bullet stops in the block and the block and bullet
combination move 2.0 m. The coefficient of kinetic friction between the block and
surface of the table is 0.2. Find the initial speed of the bullet.

Relevant Equations

$$ KE = \frac{mv^2}{2} $$
$$ W = \Delta E = F * distance $$
$$ F_friction \le F_N * \mu $$
$$ g = 10 $$

$$ \frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2) $$
v = 56.64
I just don't get how this is the wrong answer....it's just simple conservation of energy, right?

 

[erobz]

Homework Statement: A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block which is
resting on a horizontal table. The bullet stops in the block and the block and bullet
combination move 2.0 m. The coefficient of kinetic friction between the block and
surface of the table is 0.2. Find the initial speed of the bullet.
Relevant Equations: $$ KE = \frac{mv^2}{2} $$
$$ W = \Delta E = F * distance $$
$$ F_friction \le F_N * \mu $$
$$ g = 10 $$

$$ \frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2) $$
v = 56.64
I just don't get how this is the wrong answer....it's just simple conservation of energy, right?

Energy is not conserved. The bullet deforms the block and generates heat in the process. It an inelastic collision. Also, friction is doing non-conservative work on the block. Linear Momentum is conserved immediately before and after the collision.

 

[Arm]

Energy is not conserved. The bullet deforms the block and generates heat in the process. It an inelastic collision. Also, friction is doing non-conservative work on the block. Linear Momentum is conserved immediately before and after the collision.

$$ (5E-3)(v_1)=(5E-3+2)(v_2) $$
$$ \frac{1}{2} (5E-3+2)({v_2}^2)=(5E-3+2)(10)(0.2)(2) $$
$$ v_2 = 2.83 $$
$$ (5E-3)(v_1)=(5E-3+2)(2.83) $$
$$ v_1 = 1134 $$
thanks

 

[erobz]

$$ (5E-3)(v_1)=(5E-3+2)(v_2) $$
$$ \frac{1}{2} (5E-3+2)({v_2}^2)=(5E-3+2)(10)(0.2)(2) $$
$$ v_2 = 2.83 $$
$$ (5E-3)(v_1)=(5E-3+2)(2.83) $$
$$ v_1 = 1134 $$
thanks

For future reference that notation is difficult to parse. Just leave it all variables next time. I kept thinking you were taking 5 times some quantity $E$ subtracting random numbers!

 

[kuruman]

Also final numerical answers without units are meaningless. If I told you that I make two million a year, you might think I live on easy street. Not so fast. If the units of the two million is Ugandan shillings, that translates to $500.00 USD waaaay below the poverty line.