Hello!
This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).
We can write out the conservation of momentum formula:
m1vi + m2vi = m1vf + m2vf
Let:
m1 = mass of puck
m2 = mass of the goalkeeper
We know that the initial velocity of the goalkeeper is 0, so:
m1vi + m2(0) = m1vf + m2vf
m1vi = m1vf + m2vf
The final velocities will be the same, so:
m1vi = (m1 + m2)vf
Plug in the given values:
(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s
Using the equation for momentum:
p = mv
The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of
p = (120)(0.0533) = 6.396 kgm/s.
Given : Time taken to reach the maximum height t=3 s a=−g=−10m/s
2
The initial velocity of the ball can be calculated by,
Using v=u+at
∴ 0=u−10×3 ⟹u=30 m/s
Using S=ut+
2
1
at
2
∴ S=30×3+
2
1
×(−10)×3
2
=45m
Answer:
-3.5m/s²
Explanation:
Given
Initial Velocity, u = 30m/s
Final Velocity, v = 23m/s
Time, t = 2.0s
Required
Determine the magnitude of acceleration
This is calculated using first equation of motion.
v = u + at
Substitute values for v, u and t
23 = 30 + a * 2.0
23 = 30 + 2.0a
23 - 30 = 2.0a
-7 =2.0a
Solve for a
a = -7 ÷ 2.0
a = -3.5m/s²
It's A.Gas molecules moving more quickly..... hope it helps you dear
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