Let's now answer number 1:
Momentum is mass x velocity
p = mv
Block 1 is 10.00kg and is moving at a speed of 4.50m/s. Just put in what you know to get the answer:
p= 10.00kg x 4.50 m/s = 45 kg.m/s
The momentum of block 1 is 45 kg.m/s.
2. In a collision if one object bumps into another object and they both travel the same direction, the scenario would be considered as a PERFECT inelastic collision.
We can solve for unknowns using this formula:
[tex]m_{1}v_{1i}+m_{2}v_{2i} = (m_{1}+m_{2})v_{f}[/tex]
Where:
m1 = mass of object 1
m2 = mass of object 2
v1i = velocity of object 1 before collision
v2i = velocity of object 2 before collision
So before we solve for what you need, let's list down what you know based on the problem.
m1 = 10.00kg (Block1)
m2 = 25.0kg (Block2)
v1i = 4.50 m /s (Block1)
v2i = 0 m/s (Block2)
You may be wondering why the velocity of object 2 is 0m/s. Well, the problem says that Block 2 was initially at rest, then that means it is not moving, so it's speed would be 0 m/s.
Now use the formula and put in what you know and derive what you do not know.
[tex]m_{1}v_{1i}+m_{2}v_{2i} = (m_{1}+m_{2})v_{f}[/tex]
[tex](10.00kg)(4.50m/s)+(25.0kg)(0m/s) = (10.00kg + 25.00kg)v_{f}[/tex]
[tex](45kg.m/s)+(0) = (35kg)v_{f}[/tex]
[tex](45kg.m/s)= (35kg)v_{f}[/tex]
[tex] \frac{45kg.m/s}{35kg} = v_{f} [/tex]
[tex]1.29m/s = v_{f}[/tex]
Final velocity or vf of both blocks after collision is 1.29m/s.
3. To get the momentum after collision,just keep in mind that momentum is conserved even after collision. So your answer would be 45kg.m/s. If it is necessary for you to solve for it, just solve for it using the equation on the right side, which is (m1+m2)vf.
(m1+m2)vf
=(10.00kg+25.0kg)(1.29m/s)
=(35kg)(1.29m/s)
=45.15kg.m/s or 45 kg.m/s
As for the next problem, a cannon being fired is also considered a collision and this type of collision is called explosion, where the momentum before and after collision is zero. Why? Because before firing, the ball is not moving and neither is the cannon.
So now that we know this, using the formula of explosion we can solve for what we need.
[tex]m_{1}v_{1}+m_{2}v_{2}=0[/tex]
Where:
m1 = mass of object 1
m2 = mass of object 2
v1= velocity of object 1
v2 = velocity of object 2
[tex]m_{1}v_{1}+m_{2}v_{2}=0[/tex]
[tex](450kg)(v1) + (2.75kg)(23.2m/s)=0[/tex]
[tex](450kg)(v1) + (63.8kg.m/s)=0[/tex]
[tex](450kg)(v1) =0 - 63.8kg.m/s[/tex]
[tex](450kg)(v1) =- 63.8kg.m/s[/tex]
[tex]v1 = \frac{-63.8k.m/s}{450kg} [/tex]
[tex]v1 = -0.14 m/s[/tex]
The cannon travelled -0.14m/s.
Notice that the value is negative, this means that it went the opposite direction it initially traveled or it traveled backwards.
