Question

1) A 1150 kg car drives along a city street at 27.0 miles per hour (12.1 m/s). What speed must the 0.142 kg baseball have if its momentum is to be equal in magnitude to that of a car? (answer must be in m/s)

2) Two air-track carts move toward one another on an air track. Cart 1 has a mass of 0.32 kg and a speed of 1.4 m/s. Cart 2 has a mass of 0.68 kg.
   A) What speed must cart 2 have if the total momentum of the system is zero?
   B) Since the momentum of the system is zero, does it follow that the kinetic energy of the system is also zero?
   C) Verify your answer to part B by calculating the system's kinetic energy

3) A 0.150kg baseball is dropped from rest. If the magnitude f the baseball's momentum is 0.800 kg*m/s just before it lands on the ground, from what height was it dropped.

*** no work is necessary. I got these wrong on the quiz I took and would like to be able to trace my own steps bacck from the right answer. THANK YOU!

Answer 1

Question 1
========
m_car * speed_car = m_ball * speed _ball

1150x12.1= 0.142x speed_ball

speed_ball= 97992.95 m/s

Ans- Speed of the ball must be 97992.95 m/s

Qs 2 )

Data:
m1:=0.32 kg; m2:=0.68 kg; v1:=1.4 m/s;

Total momentum equation:
p = m1*v1 + m2*v2

Set p=0 and solve for v2:
v2 = -m1*v1/m2

= - 0.32x 1.4/ 0.68

= -0.6588 m/s (Negative sign indicates that its velocity is in the direction opposite that of cart 1. )

Ans- a) cart 2 should have 0.6588 m/s speed if the total momentum of the system is zero

b) NO. Momentum adds up as vectors depending on directions of velocity. Kinetic energy is a scalar, and is always positive. Adding up kinetic energy will result in a number greater than individual kinetic energies.

c) And the kinetic energy:
KE = 1/2*m1*v1^2 + 1/2*m2*v2^2

KE = 1/2*m1*v1^2 + 1/2*m2*(-m1*v1/m2)^2
KE = 1/2*v1^2*(m1 + m1^2/m2)

= 0.5x 1.4x1.4(0.32+0.32^2/0.68)

=0.98x0.47

Ans- c) Systems kinetic energy is =0.461 J

Ques 3)

Use m_ball * vf = momentum
0.15 * vf = 0.80 kg m/s
vf = 5.33 m/s

vi = 0
vf = 5.33 m/s
g = 9.8m/s^2
d = ???

Use vf^2 = vi^2 + 2*g*d
5.33^2=2x9.8xd
d = 1.45 m

Ans- Height is 1.45 m