Membangun Indonesia dengan Fisika: Problem and Solutions on Impulse Momentum

Senin, 07 November 2011

Problem and Solutions on Impulse Momentum

Impulse Momentum Problem and Solutions

1. An object travels with a velocity 4m/s to the east. Then, its direction of motion and magnitude of velocity are changed. Picture given below shows the directions and magnitudes of velocities. Find the impulse given to this object.

I=F.Δt=Δp=m.ΔV

where ΔV=V₂-V₁=-3-4=-7m/s

I=m.ΔV=3.(-7)=-21kg.m/s

2. Ball having mass 4kg and velocity 8m/s travels to the east. Impulse given at point O, makes it change direction to north with velocity 6m/s. Find the given impulse and change in the momentum.

Initial and final momentum vectors of ball are shown in the figure below.

P₁=m.V₁=4kg.8m/s=32kg.m/s

P₂=m.V₂=4kg.6m/s=24kg.m/s

ΔP=P₂+P₁ (vector addition)

ΔP²=P₂²+P₁²=m²(v₂²+v₁²)

ΔP²=16.100

ΔP=40kg.m/s

Impulse=change in momentum

I=ΔP=40kg.m/s

3. Find the impulse and force which make 12m/s change in the velocity of object having 16kg mass in 4 s.

F.Δt=ΔP=m.ΔV

F.4s=16kg.12m/s

F=48N

F.Δt=Impulse=192kg.m/s

4. Applied force vs. time graph of object is given below. Find the impulse of the object between 0-10s.

Area under the force vs. time graph gives us impulse.

F.Δt=20.2/2+20.(6-2)+20.(10-6)/2

F.Δt=140kg.m/s

5. A ball having mass 500g hits wall with a10m/s velocity. Wall applies 4000 N force to the ball and it turns back with 8m/s velocity. Find the time of ball-wall contact.

F.Δt=ΔP=m.ΔV=m.(V₂-V₁)

-4000.Δt=0,5kg.(8-10)

Δt=0,00025s

6. Objects shown in the figure collide and stick and move together. Find final velocity objects.

Using conservation of momentum law;

m₁.V₁+m₂.V₂=(m₁+m₂).V_final

3.8+4.10=7.V_final

64=7.V_final

V_final=9,14m/s

7. 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.

a) Using conservation of momentum law;
(m₁+m₂).V=m₁.V₁+m₂.V₂

5.4=30+2.V₁

V₁=-5m/s

b) E_Kinitial=1/2/m₁+m₂).V²

E_Kinitial=1/2.5.16=40joule

E_Kfinal=1/2.2.5²+1/2.3.10²

E_Kfinal=175 joule

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

8. As shown in the figure below, object m₁ collides stationary object m₂. Find the magnitudes of velocities of the objects after collision. (elastic collision)

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

9. Momentum vs. time graph of object is given below. Find forces applied on object for each interval.

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

9. 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.