PLEASE SHOW FULL STEPS: problem 1: A projectile is ?red into the air from ground

Question

PLEASE SHOW FULL STEPS:

problem 1: A projectile is ?red into the air from ground level over level ground. The initial speed of the projectile is 30 m/s and the launch angle of the projectile is 45?. At the highest point in its ?ight the projectile explodes into two pieces of equal mass. Immediately after the explosion one piece has a momentary speed of zero and then falls straight to the ground. Neglect air resistance.

A) Where does the center of mass of the projectile land?

B) Where does the second piece land?

C) What is the speed of the second piece immediately after the explosion?

problem 2: Consider a 1-dimensional collision between blocks of objects of m1 = 2 kg and m2 = 5 kg. The collision occurs along the x-direction Before the collision m1 has an x-component of velocity of 10m/s and m2 has an x-component of velocity of ?6 m/s.

A) Find the x-component of velocity of m2.    

B) Find Q for this collision. Recall Ktot,bef - Ktot,aft = Q   

problem 3: Consider a 1-dimensional collision between blocks of objects of m1 = 2 kg and m2 = 5 kg. The collision occurs along the x-direction Before the collision m1 has an x-component of velocity of 10m/s and m2 has an x-component of velocity of ?6 m/s.

A) If the objects stick together as a result of the collision, find the x-component of velocity of the two blocks after the collision.    

B) If the objects stick together as a result of the collision, find the difference between the total kinetic energy of the blocks before the collision and the total kinetic energy of the blocks after the collision.. That is find Q, where Ktot,bef=_Ktot,aft + Q

Explanation / Answer

problem 2:

A)

v2f = (2*2)/(2+5) * (10) + (5-2)/(2+5) * (-6)

v2f = 3.1429

v2f = 3.14 m/s

B)

v1f = (2-5)/(2+5) * (10) + (2*5)/(2+5) * (-6)

v1f = -12.857 m/s

K_b = 0.5 m1 v1i^2 + 0.5 m2 v2i^2 = 0.5*2*10*10 + 0.5*5*6*6 = 190 J

K_a = 0.5 m1 v1f^2 + 0.5 m2 v2f^2 = 0.5*2*12.857*12.857 + 0.5*5*3.1429*3.1429 = 190 J

==> Q = 0

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problem 3:

A)

m1 v1 + m2 v2 = (m1 + m2) (v)

2*10 + 5*(-6) = (2+5) * v

==> v = -1.4286

==> v = -1.43 m/s

B)

K_b = 0.5 m1 v1i^2 + 0.5 m2 v2i^2 = 0.5*2*10*10 + 0.5*5*6*6 = 190 J

K_a = 0.5 (m1+m2) v^2 = 0.5 * (2+5) * (1.4286*1.4286) = 7.1431 J

Q = 190 - 7.1431 = 182.8569 = 183 J

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