Answer:
18. They will travel to the right,
19. They will travel to the left.
Explanation:
Momentum
Recall that the direction of a collision depends on the sign of the system's net momentum or the sum of all momentums. (\(\star\))
Also recall that p = mv where m is the mass of an object and v is its velocity.
Initially, the system of car A and car B has an initial net momentum of
\(m_Av_A,_i+m_Bv_B,_i\),
where \(v_a,_i \:and\:v_B,_i\) are the initial velocities of each car.
If the directions up/right are considered positive and down/left negative, then car B's momentum has a negative value, making \(p_i\) be,
\(m_Av_A,_i-m_Bv_B,_i\).
Applying Momentum to Problem 18
If problem 18 says car A's momentum is larger, then that means the difference of car A's and B's momentum produces a larger positive value, thus the car moves in the positive direction or to the right!
Plugging in various values of car A's and B's momentum can also be used to come to this conclusion.
\(p_A < p_B:\)
Let, car A's momentum have a value of 1 and car B's has a value of 3, then there net momentum initially is 1 - 3 or -2 (negative direction).
\(p_A > p_B:\)
Now, let car A's momentum have a value of 10 and car B's value stays the same, then the net momentum is 10 - 3 or 7 (positive direction).
Applying Momentum to Problem 19
If car A and B have the same m and B has a larger velocity. Then car B has a larger momentum. This makes the net momentum or \(\rm p_i\) negative thus, moving the vehicles to the left.
We can also plug in values of m and v to confirm this.
\(m_B > m_A:\)
Let v = 3, \(m_A=4\) and \(m_B=8\),
\(p_i=4(3)-8(3)=-12\).
