The mass of first ball = m1 = 50 g = 0.05 kg, mass of the second ball = m2- 100 g = 0.1 kg Initial velocity of the first ball = u1 = 3 m/s, Initial velocity of the second ball = U2 = 1.5 m/s Final velocity of the second ball = V1= ? Final velocity of first ball
Answers
Explanation:
The linear momentum of a particle with mass m moving with velocity v is defined as
p = mv (7.1)
Linear momentum is a vector . When giving the linear momentum of a particle you must
specify its magnitude and direction. We can see from the definition that its units must be
kg·m
s
. Oddly enough, this combination of SI units does not have a commonly–used named so
we leave it as kg·m
s
!
The momentum of a particle is related to the net force on that particle in a simple way;
since the mass of a particle remains constant, if we take the time derivative of a particle’s
momentum we find
dp
dt = m
dv
dt = ma = Fnet
so that
Fnet =
dp
dt (7.2)
7.1.2 Impulse, Average Force
When a particle moves freely then interacts with another system for a (brief) period and
then moves freely again, it has a definite change in momentum; we define this change as the
impulse I of the interaction forces:
I = pf − pi = ∆p
Impulse is a vector and has the same units as momentum.
When we integrate Eq. 7.2 we can show:
I =
Z tf
ti
F dt = ∆p
Answer:
Using Momentum Conservation:
⇒m
1
u
1
+m
2
u
2
=m
1
v
1
+m
2
v
2
⇒1(21)+2(−4)=1(1)+2v
2
⇒v
2
=6m/s
Coefficient of Restitution:
e=
u
1
−u
2
v
2
−v
1
=
21+4
6−1
=0.2
Kinetic Energy lost:
=
2
1
m
1
(v
1
2
−u
1
2
)+
2
1
m
2
(v
2
2
−u
2
2
)
=
2
1
1(1−441)+
2
1
2(36−16)
=−200J
Impulse of Force:
I=(Δmv)
1
=1(1−21)=−20
=(Δmv)
2
=2(6−(−4))=20