Impulse Momentum Exam 2 and Problem Solutions

1. Objects shown in the figure collide and stick and move together. Find final velocity objects. Impulse Momentum Using conservation of momentum law;

m₁.V₁+m₂.V₂=(m₁+m₁).Vfinal

3.8+4.10=7.Vfinal

64=7.Vfinal

Vfinal=9,14m/s

2. 2kg and 3kg objects slide together, and then they break apart. If the final velocity of m₂ is 10 m/s,

a) Find the velocity of object m₁.

b) Find the total change in the kinetic energies of the objects. Impulse Momentum a) Using conservation of momentum law;

(m₁+m₂).V=m₁.V₁+m₂.V₂

5.4=30+2.V₁

V₁=-5m/s

b) EKinitial=1/2/m₁+m₂).V₂

EKinitial=1/2.5.16=40joule

EKfinal=1/2.2.52+1/2.3.102

EKfinal=175 joule

Change in the kinetic energy is =175-40=135 joule

3. As shown in the figure below, object m₁ collides stationary object m₂. Find the magnitudes of velocities of the objects after collision. (elastic collision) Impulse Momentum In elastic collisions we find velocities of objects after collision with following formulas;

V₁’=(m₁-m₂)/(m₁+m₂).V₁

V₂’=(2m₁/m₁+m₂).V₁

m₁=6kg, m₂=4kg, V₁=10m/s

V₁’=(6-4/6+4).10=2m/s

V₂’=(2.6/6+4).10=12m/s

4. Momentum vs. time graph of object is given below. Find forces applied on object for each interval. Impulse Momentum F.Δt=ΔP

F=ΔP/Δt

Slope of the graph gives us applied force.

I. Interval:

F₁=P₂-P₁/10-0=-50/10=-5N

II. Interval:

F₂=50-50/10=0

III. Interval:

F₃=100-50/10=5N

5. A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m₁, collide box having mass m₂ and they move together. Find the velocity of boxes. Impulse Momentum Energy stored in the spring is transferred to the object m1.

1/2.k.X²=1/2.mV²

50N/m.(0,2)²=0,5.V²

V=2m/s

Two object do inelastic collision.

m₁.V₁=(m₁+m₂).Vfinal

0,5.2=2.Vfinal

Vfinal=0,5m/s