Question

Question 8 (3 points)

Bob throws a sticky piece of gum at a stationary cardboard box. The gum has a mass of 0.003 kg.
and the box has a mass of 0.05 kg. The gum hits the box with a speed of 9 m/s, and they stick
together and slide forward.
a. What is the total momentum of the system before the collision? (1 point)
b. What is the total momentum of the system after the collision? (1 point)
c. What is the velocity of the gum/box combination after the collision? (1 point)

Answer 1

p=mv

a).p=(0.003kg)(9m/s)+(0.05kg)(0m/s)=0.027kg*m/s

Answer 2

a) 0.027 kgm/s

b) 0.027 kgm/s

c) 0.509 m/s

Step-by-step explanation:

Mass of gum = m1 = 0.003 kg

Mass of box = m2 = 0.05 kg

Velocity of gum before collision = V1 = 9 m/s

Velocity of box before collision = V2 = 0 m/s

a) Momentum of system before collision = m1V1 – m2V2

= (0.003)(9) – (0.05)(0) = 0.027 kgm/s

b)According to law of conservation of momentum,

Momentum of system after collision = Momentum of system before collision

Momentum of system after collision = 0.027 kg/s

c) as the gum and box behaving as a single object so their velocity (V’) will be same.

Momentum of system after collision = (m1 + m2)V’ = 0.027 kg/s

V’ = 0.027/(m1 + m2)

V’ = 0.027/(0.003 + 0.05)

V’ = 0.509 m/s