Complete question:
Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear (Fig. 7–43). Car A has a mass of 435 kg and car B 495 kg, owing to differences in passenger mass. If car A approaches at 4.50 m/s and car B is moving at 3.70 m/s, calculate (a) their velocities after the collision, and (b) the change in momentum of each.
(Check image uploaded for Fig. 7–43)
Answer:
(a) velocity of car A after collision is 1.637 m/s
velocity of car B after collision is 2.437 m/s
(b) change in momentum of car A is 1245.41 kg.m/s
change in momentum of car B is 625.19 kg.m/s
Explanation:
Given;
mass of car A , [tex]M_A[/tex] = 435 kg
mass of car B, [tex]M_B[/tex] = 495 kg
velocity of car A before collision, [tex]U_A[/tex] = 4.5 m/s
velocity of car B before collision, [tex]U_B[/tex] = 3.7 m/s
The collision between the two cars is elastic, then we consider the principle of conservation of linear momentum;
[tex]M_AU_A +M_BU_B = M_AV_A +M_BV_B[/tex]
where;
[tex]V_A \ is \ the \ velocity \ of \ car \ A \ after \ collision \\\\V_B \ is \ the \ velocity \ of \ car \ B\ after \ collision[/tex]
[tex]435*4.5 +495*3.7 = 435V_A +495V_B\\\\3789 = 930(V_A+V_B)\\\\V_A+V_B = \frac{3789}{930} = 4.074\\\\V_A+V_B = 4.074 -------equation (i)[/tex]
Also, since the collision is elastic, the velocity of car B after collision will be;
[tex]V_B = U_A-U_B + V_A[/tex]
Substitute for initial velocity, then we will have;
[tex]V_B = 4.5-3.7 + V_A\\\\V_B =0.8 +V_A[/tex]
Substitute VB in equation (i)
[tex]V_A+0.8+V_A = 4.074 \\\\2V_A = 4.074 - 0.8\\\\2V_A = 3.274\\\\V_A = \frac{3.274}{2} = 1.637\ m/s[/tex]
[tex]V_B = 0.8+1.637\\\\V_B = 2.437 \ m/s[/tex]
Part (b) Change in momentum, ΔP = ΔMV
For car A
ΔP = ΔMV = MU - MV
= M(U - V)
= 435 (4.5 - 1.637) = 1245.41 kg.m/s
For car B
ΔP = ΔMV = MU - MV
= M(U - V)
= 495 (3.7 - 2.437) = 625.19 kg.m/s
The image in the question is missing and i have attached it here.
Answer:
A) Final Velocity of Car A = 3.648 m/s
Final velocity of Car B = 4.448 m/s
B) Change in momentum of Car A = 370.62Kg.m/s
Change in momentum of Car B = 370.26 Kg.m/s
Explanation:
A) Mass of car A is 435 kg.
Mass of car B is 495 kg.
Initial speed of car A is 4.5 m/sec.
Initial speed of car B is 3.7 m/sec.
From law of conservation of energy,
M1U1 + M2U1 =M1V1 + M2V2
So in this case;
(435 x 4.5) + (495 x 3.7) = 435V1 + 495V2
And so;
3789 = 435V1 + 495V2 ------- eq 1
We also know that the relative velocity of one particle with respect to the other is reversed by the collision. Thus;
U1 - U2 = - (V1 - V2)
And so, U1 - U2 = V2 - V1
Thus, V2 - V1 = 4.5 - 3.7 = 0.8 m/s
V2 = V1 + 0.8
Putting V1 + 0.8 for V2 in eq 1 to obtain;
3789 = 435V1 + 495(V1 + 0.8)
3789 = 435V1 + 495V1 + 396
3789 - 396 = 930 V1
V1 = 3393/930 = 3.648 m/s
Since V2 = V1 + 0.8
Thus; V2 = 3.648 + 0.8 = 4.448 m/s
B) Momentum of Car A before collision = M1U1 = 435 x 4.5 = 1957.5 Kg.m/s
Momentum of car A after collision is given as;
= M1V1 = 435 x 3.648 = 1586.88 Kg.m/s
we know that change in momentum is given by;
Change in momentum = M1U1 - M1V1
So change in momentum of Car A =
1957.5 - 1586.88 = 370.62Kg.m/s
Similarly;
Momentum of Car B before collision = M2U2 = 495 x 3.7 = 1831.5 Kg.m/s
Momentum of car B after collision is given as;
= M2V2 = 495 x 4.448 = 2201. 76 Kg.m/s
we know that change in momentum is given by;
Change in momentum = M2V2 - M2U2
So change in momentum of Car B=
2201.76 - 1831.5 = 370.26 Kg.m/s
