Total mechanical energy = kinetic energy + potential energy
E = KE + PE
E = ½mv² + mgh
E = ½(0.1 kg)(2 m/s)² + (0.1 kg)(9.8 m/s²)(1.5 m)
E = 0.2 J + 1.47 J
E = 1.67 J
Answer:
v1 = 15.90 m/s
v2 = 8.46 m/s
mechanical energy before collision = 32.4 J
mechanical energy after collision = 32.433 J
Explanation:
given data
mass m = 0.2 kg
speed = 18 m/s
angle = 28°
to find out
final velocity and mechanical energy both before and after the collision
solution
we know that conservation of momentum remain same so in x direction
mv = mv1 cosθ + mv2cosθ
put here value
0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)
3.6 = 0.1765 V1 + 0.09389 v2 ................1
and
in y axis
mv = mv1 sinθ - mv2sinθ
0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)
0 = 0.09389 v1 - 0.1768 v2 .......................2
from equation 1 and 2
v1 = 15.90 m/s
v2 = 8.46 m/s
so
mechanical energy before collision = 1/2 mv1² + 1/2 mv2²
mechanical energy before collision = 1/2 (0.2)(18)² + 0
mechanical energy before collision = 32.4 J
and
mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²
mechanical energy after collision = 32.433 J
Rocket fuel will and smoke will emit from the thrusters.
if there is not a reaction
The unit is light years or Ly
