Download Impulse Momentum (Problem and Solutions) 1. An object travels

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
1. 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).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.
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)
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.
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 m 1, collide box
having mass m2 and they move together. Find the velocity of boxes.
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