A 21g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging

Question

A 21g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s. What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move? 0 J 9.9 kJ -9.9 kJ 10.2 J -86.7J Neglecting gravity, doubling the exhaust velocity from a single stage rocket initially at rest changes the final velocity attainable by what factor? Assume all other variables, such as the mass of the rocket and the mass of the fuel, do not change. The final velocity stays the same. The final velocity doubles. The final velocity increases by a factor of 0.693. The final velocity increases by a factor of 0.310. A 6-kg object is moving to the right at 4 m/s and collides with a 4-kg object moving to the left at 6 m/s. The objects collide and stick together. After the collision, the combined object: has the same kinetic energy that the system had before the collision. has more kinetic energy than the system had before the collision. has no kinetic energy has less momentum than the system had before the collision.

Explanation / Answer

1. We use the conservation of momentum theory in this case considering the collision

Therefore, the theory states that mu=(M+m)v

where u=1000m/s

and v=200m/s

M and m are the mass of the block and bullet

initial kinetic energy =1/2mu^2=.5*.021*1000^2=10500J

Final kinetic energy =1/2(M+m)v^2=.5(1+.021)*200^2=20420J

difference in kinetic energy=initial-final=10500-20420=-9920J=-9.92KJ

2) We are considering two conditions:

the single stage rocket and its initial rest condition

If the exhaust velocity is Ve then the final velocity considering both condiions and neglecting gravity and air resistance is Vf=Ve.ln(M0/Mr)

where M0=initial rocket mass and Mr=mass after exhaustion

according to the question M0/Mr does not change

So ln(M0/Mr) is constant

That means Vf is directly dependent on Ve

so if Ve becomes 2Ve then Vf becomes 2Vf

therefore the final velocity doubles

3) According to the conservation of momentum

m1u1+m2u2=(m1+m2)v

or 6.4+4*6=(6+4)v

or 48/10=v

or v=4.8m/s

initial kinetic energy=1/2m1u1^2+1/2m2u2^2=48+72=120J

final kinetic energy =1/2(m1+m2)v^2=115.2J

has less kinetic energy before the collision

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