# Ch 7.1 Help ## Ch 7.1 Help

Question 1
An object of 1.1988 kg mass has a horizontal velocity component of 2.409 m/s and a vertical velocity component of 3.82 m/s. What is the magnitude of the momentum of the particle (see sheet 3; (7.1) stands for 2 equations, x-and y-component)? Indicate with a negative (positive) sign whether the SI unit of momentum is the combination kg·m/s reflecting the definition of momentum (is the combination kg·m/s contracted into a new name).

p = √((m*v1)2+(m*v2)2))
p = √((1.1988*2.409)2+(1.1988*3.82)2))
p = √(2.88790922+4.5794162)
p = √(8.340019547+20.9710509)
p = √(29.31107045)
p =
Spoiler:
5.413969934

Question 2
An object is acted upon by a force of 11.8 N for 0.1342 s. What is the change in the object's momentum (see sheet 4)? Indicate with a negative (positive) sign whether the direction of the momentum-change vector is in the direction (opposite direction) of the force?

M = F*t
M = 11.8*0.1342
M =
Spoiler:
1.5836

Question 3
A 139.2 gram object with an initial velocity of 4.418 m/s is brought to rest by a 17.23 N force. How long is the force active (see sheet 3,4)?

F(t) = (v/F)*mkg
F(t) = (4.418/17.23)*0.1392 (Remember to change the mass from grams to kiligrams)
F(t) = (0.2564132327)*0.1392
F(t) =
Spoiler:
0.035692722

Question 4
Two cars collide in a perfectly inelastic collision. The mass of car(1) is 2,409 kg and the mass of car(2) is 1,706 kg. What is the velocity of car(2) after the collision if the velocity of car(1) after the collision is 21.19 m/s?

vcar 2 = vcollision
vcar 2 =
Spoiler:
21.19

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kristy
Guest

## Question on 4

Why does the velocity of the second car equal the velocity of the collision? I tried doing the conservation of momentum equation, but clearly that wasn't the right way to go. Could you please explain?

S
Guest

I love you too

physics
Guest

## Re: Ch 7.1 Help

I believe it is because an object in motion stays in motion, and when that object collides with another object the energy is transferred to the second object. So if a ball is thrown at 60mph and hits a second ball that second will continue to travel at 60mph. I think it's supposed to be implied that this all happens in a vacuum, or at least without any resistance. This is also for an inelastic collision, so energy is directly transferred. Perhaps the idea of Billiards will help, when the cue ball strikes a numbered ball, that numbered ball will continue at the same speed. Take that example as it is, don't over think it.

If someone knows a correct, or more correct answer, please post. This is to the best of my understanding.

I love you all.

## Re: Ch 7.1 Help

Yeah, the textbook states that in a perfectly inelastic collision that the two objects stick together and because of this their final velocities must be the same.

guesto
Guest

## Re: Ch 7.1 Help 