Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
0.278 m/s
Explanation:
We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.
So we can write:
where
m = 0.200 kg is the mass of the koala bear
u = 0.750 m/s is the initial velocity of the koala bear
M = 0.350 kg is the mass of the other clay model
v is their final combined velocity
Solving the equation for v, we get