The law of conservation of momentum states that for a system of objects on which

ID: 2033091 • Letter: T

Question

The law of conservation of momentum states that for a system of objects on which there is no net external force, the total momentum remains constant. The aim of this laboratory is to investigate this for collisions between two carts as illustrated in Pig. 1. Before collision After collision Figure 1: Colliding carts The momentum of any single cart at a given instant is p = mv (p and v are vectors) where m is the mass of the cart and v is the velocity of the cart at the instant. The total momentum of the system at any instant is protal Pa Pawhere pais the momentum of cart A at that instant and pa is the momentum of cart B at the same instant. Conservation of momentum implies that Potal beforePhotalatern This laboratory exercise aims to verify this fact 6 Exercises a) A cart of mass mi moving to the right collides with a cart of mass m2 that is at rest. If m2is significantly larger than mi,consider the following two different inelastic collisions (the amount of energy lost is different in each and does not matter in the problem), then several scenarios are possible. In one scenario the cart on the left rebounds as illustrated in Fig. 5. Before collision After collision Rest Figure 5 Rebound

Explanation / Answer

a. given

initial speed of m1 = u1

for scenario 1

final speed of m1 = v1

intiail speed of m2 = 0

final speed of m2 = v2

from conservation foe mmeomentum

m1u1 = -m1v1 + m2v2

v2 = (m1)(u1 + v1))/m2

for 2nd scenario

v1 = 0

hence

v2 = (m1)u1/m2

hence the car2 moves faster in scenario 1

b. m2 = 0.5 m1

after the plunger triggers the carts, it plunges them with same force, and hence acceleration of each cart is inversly proporitonal to mass

now, the force actrs for same duraiotn on both masses

hence

acceleration of m2 > m1, because m2 < m1

hence

m2 moves with greater speed

c. m2 = 0.5 m1

u = 4 m/s

after the plunger triggers

v2 = 6 m/s

from conservation of moemntum

m1v1 + m2v2 = (m1 + m2)u

v1 + v2 = 1.5u

v2 = 1.5u - v1 = 6-6 = 0

hence v1 = 0

hence the 1st block comes to rest

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The law of conservation of momentum states that for a system of objects on which

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