Answer:
a). 1.218 m/s
b). R=2.8
Explanation:


Momentum of the motion the first part of the motion have a momentum that is:


The final momentum is the motion before the action so:
a).




b).
kinetic energy

Kinetic energy after

Kinetic energy before

Ratio =

Answer:
80 m/s
Explanation:
Given:
a = -5 m/s²
v = 0 m/s
Δx = 640 m
Find: v₀
v² = v₀² + 2a(x − x₀)
(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)
v₀ = 80 m/s
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart, 
Initial speed of the cart, 
Mass of the larger cart, 
Initial speed of the larger cart, 
After the collision,
Final speed of the smaller cart,
(as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let
is the speed of the large cart after collision. It can be calculated using conservation of momentum as :





So, the speed of the large cart after collision is 0.301 m/s.
Answer:
a) 69.3 m/s
b) 18.84 s
Explanation:
Let the initial velocity = u
The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively
uᵧ = u sin 40° = 0.6428 u
uₓ = u cos 40° = 0.766 u
We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m
The range of a projectile motion is given as
R = uₓt
where t = total time of flight
1000 = 0.766 ut
ut = 1305.5
The vertical distance travelled by the projectile rocks,
y = uᵧ t - (1/2)gt²
y = - 900 m (900 m below the crater's level)
-900 = 0.6428 ut - 4.9t²
Recall, ut = 1305.5
-900 = 0.6428(1305.5) - 4.9 t²
4.9t² = 839.1754 + 900
4.9t² = 1739.1754
t = 18.84 s
Recall again, ut = 1305.5
u = 1305.5/18.84 = 69.3 m/s
To solve this problem, we are going to use the formula for
work which is Fd where x and y are measured separately.
X direction: W = 13.5 x 230 = 3105 Joules
Y direction: W = -14.3 x -165 = 2360 Joules
So the total work is getting the sum of the two: 3105 + 2360
= 5465 Joules