(a) The momentum of the first trolley is 5.4 kgm/s
(b) The velocity of the trolleys after impact is 2.7m/s
<u>Explanation:</u>
Given:
Mass, m₁ = 1.2kg
Velocity, v₁ = 4.5m/s
Mass, m₂ = 0.8kg
v₂ = 0
(a) Momentum of the trolley before impact, p
We know:
Momentum = mass X velocity
p = 1.2 X 4.5
p = 5.4 kgm/s
Therefore, the momentum of the first trolley is 5.4 kgm/s
(b) Speed of the trolleys after impact, v = ?
During collision, the momentum is conserved.
So,
m₁v₁ + m₂v₂ = (m₁ + m₂)v
(1.2 X 4.5) + (0.8 X 0) = (1.2 +0.8) X v
5.4 + 0 = 2v
v = 2.7m/s
Therefore, the velocity of the trolleys after impact is 2.7m/s
Answer:
The time taken is 
Explanation:
From the question we are told that
The speed is 
The length of the tennis court 
Generally the time taken is mathematically represented as

=> 
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
Answer:
t=0.42s
Explanation:
Here you have an inelastic collision. By the conservation of the momentum you have:

m1: mass of the bullet
m2: wooden block mass
v1: velocity of the bullet
v2: velocity of the wooden block
v: velocity of bullet and wooden block after the collision.
By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

hence, the time is t=0.42 s