1.) A missile of mass 1.25 102 kg is fired from a plane of mass 4.65 103 kg init
ID: 1965804 • Letter: 1
Question
1.) A missile of mass 1.25 102 kg is fired from a plane of mass 4.65 103 kg initially moving at a speed of 3.50 102 m/s. If the speed of the missile relative to the plane is 1.11 103 m/s, what is the final velocity of the plane?2.) A bullet of mass 10.7 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.02 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 4.60 m/s.
(a) Find the initial speed of the bullet.
(b) How much kinetic energy is lost?
Explanation / Answer
This is answer for 2)
Bullet:
M1=10.7 g=10.7*10^-3 Kg
Va1=?
Va2= Vb2=4.6 m/s
Block:
M2=1.02 Kg
Vb1=0
Vb2=4.60 m/s
A)
In collision, Momentum before collision=momentum after collision.
Therefore, M1Va1+M2Vb1=M1Va2+M2Vb2
Va1=(M1Va2+M2Vb2-M2Vb1)/M1
Plug in and find the initial velocity of Bullet.
B)
Kinetic energy before collision was only of bullet as the block was at rest.
K.E1=(1/2)M1*Va1^2
The bullet embeds in the block after collision, making the mass to be M1+M2. Therefore kinetic energy after collision is:
K.E2=(1/2)(M1+M2)*Vb2^2
Kinetic energy lost=Difference in K.E1 and K.E2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.