A block of mass 7.33 kg in outerspace is moving at 1.72 m/s with no external forces acting on it. After an explosion, the block is split into two parts both having mass equal to half the mass of the original block. The explosion supplies the two masses with an additional 16.7 J of kinetic energy. Neither mass leaves the line of original motion. Calculate the magnitude of the velocity of the mass that is moving at a greater velocity.
The original momentum is mv
the final momentum is m/2 * v1' + m/2*v2'
set the equal.
2v=v1' + v2'
a) v1'=2v-v2'
Now, energy.
1/2 m v^2+16.7=1/4 m v1'^2 + 1/4 m v2'^2
b) 2 v^2+ 16.7=v1'^2 + v2'^2
put the expression a) into b). multiply it out, gather terms, and solve the quadratic.
bob- thanks, this helped alot. Just one thing though- when you're elminating the mass and the fractions in the Kinetic energy equation, you forgot about the normal number(energy added). It should be 4 * (16.7) / m after doing the elimination. The rest was perfect though, thanks!
wow thanks
I don't understand why 16.7 J of kinetic energy has to be added to the initial kinetic energy of the mass instead of the final kinetic energy of the two split masses. Anyone have a good explanation ?
To solve this problem, we need to apply the principle of conservation of momentum and conservation of kinetic energy.
Let's first find the initial momentum of the block before the explosion:
Initial momentum (before explosion) = mass × velocity
p_initial = 7.33 kg × 1.72 m/s
p_initial = 12.6036 kg·m/s
The explosion splits the block into two parts, with each part having a mass of half the original mass (3.665 kg). Let's assume the velocities of the two parts are v1 and v2, with v1 being the greater velocity.
According to conservation of momentum, the total momentum after the explosion is equal to the initial momentum before the explosion. Therefore:
p_initial = p1 + p2
Since the direction of motion is the same, the magnitude of momentum is equal to the product of mass and velocity. So we can rewrite the equation as:
mass × velocity = mass1 × velocity1 + mass2 × velocity2
Using the given information, we have:
7.33 kg × 1.72 m/s = 3.665 kg × v1 + 3.665 kg × v2
12.6036 kg·m/s = 3.665 kg × (v1 + v2)
Now, let's consider the conservation of kinetic energy. The explosion supplies the masses with an additional 16.7 J of kinetic energy. Therefore, the change in kinetic energy is equal to the supplied energy:
change in kinetic energy = 16.7 J
change in kinetic energy = final kinetic energy - initial kinetic energy
The initial kinetic energy is given by:
initial kinetic energy = (1/2) × mass × velocity^2
initial kinetic energy = (1/2) × 7.33 kg × (1.72 m/s)^2
The final kinetic energy is given by:
final kinetic energy = (1/2) × mass1 × velocity1^2 + (1/2) × mass2 × velocity2^2
final kinetic energy = (1/2) × 3.665 kg × v1^2 + (1/2) × 3.665 kg × v2^2
Plugging in the given values, we have:
(1/2) × 7.33 kg × (1.72 m/s)^2 + 16.7 J = (1/2) × 3.665 kg × v1^2 + (1/2) × 3.665 kg × v2^2
Now, we have two equations:
12.6036 kg·m/s = 3.665 kg × (v1 + v2)
(1/2) × 7.33 kg × (1.72 m/s)^2 + 16.7 J = (1/2) × 3.665 kg × v1^2 + (1/2) × 3.665 kg × v2^2
We can solve this system of equations to find the values of v1 and v2.