Question
1.
A ball of
mass 0.5 kg moving at 10 ms^{1} collides with another ball of equal
mass at rest. If the two balls move off together after the impact, calculate
their common velocity.
Solution:
From the
equation based on the law of conservation of momentum, we have
M_{1}u_{1}
+ m_{2}u_{2} = m_{1}v + m_{2}v
Given
m_{1} = 0.5 kg, u_{1} = 10 ms^{1}, m_{2} = 0.5
kg, u_{2} = 0, V =?
Therefore,
0.5 x
10 + 0.5 x 0 = 0.5v + 0.5v
5 + 0 = v
The common
velocity the two balls move off together with is 5 ms^{1}
Question
2.
A body of
mass 4.2 kg moving with velocity 10 ms^{1} due east, hits a stationary
body of mass 2.8 kg. If they stick together after collision and move with
velocity v due east, calculate the value of v.
Solution:
The law of
conservation of momentum equation is
M_{1}u_{1}
+ m_{2}u_{2} = m_{1}v + m_{2}v
4.2 x
10 + 2.8 x 0 = (4.2 + 2.8) v
42 +
0 = 7v
v = 42/7 = 6
The final
and common velocity the two bodies move with after collision is 6.0 ms^{1}
Question
3.
A body of
mass 5 kg moving with a velocity of 10 ms^{2} collides with a
stationary body of mass 6 kg. If the two bodies stick together and move in the
same direction after the collision, calculate their common velocity.
Solution:
The equation
from the law of conservation of momentum is
M_{1}u_{1}
+ m_{2}u_{2} = m_{1}v + m_{2}v
5
x 10 + 6 x 0 = 5v + 6v
50 + 0
= 11v
v =
50/11
= 4.55 ms^{1}
Question
4.
A tractor of
mass 5.0 x 10^{3} kg is used to tow a car of mass 2.5 x 10^{3}
kg. The tractor moved with a speed of 3.0 ms^{1} just before the towing
rope becomes taut. Calculate
(i)
speed of the tractor immediately the
rope becomes taut;
(ii)
loss in kinetic energy of the system
just after the car has started moving.
Solution:
(i). By the
law of conservation of momentum, immediately the rope becomes taut, the momentum
of the system is conserved and expressed as
m_{1}u_{1}
+ m_{2}u_{2} = m_{1}v + m_{2}v
Where m_{1}
= mass of the tractor, u_{1} = initial velocity of the tractor, m_{2}
= mass of the car, u_{2} = initial velocity of the car, v = common and
final velocity of both the tractor and the car.
Note:
since the car was initially at rest, u_{2} = 0, therefore, the initial
momentum of the car m_{2}u_{2} = 0
Therefore,
from the conservation of momentum equation,
m_{1}u_{1}
+ 0 = v (m_{1} + m_{2})
5 x 10^{3}
x 3 = v (5 + 2.5) 10^{3}
15 = v
x 7.5
v =
15/7.5
= 2
The final
common velocity is also the speed of the tractor when the rope becomes taut = 2
ms^{1}
(ii). Loss
in kinetic energy of the system equals
initial
kinetic energy of the system  final kinetic energy
i.e.
K.E_{1}  K.E_{2}
K.E_{1}
= initial kinetic energy = initial kinetic energy of the tractor + initial
kinetic energy of the car
=
K.E_{1} (tractor) + K.E_{1} (car)
Since K.E_{1}
(car) = 0, the initial kinetic energy of the system is K.E_{1}
(tractor).
Generally,
the kinetic energy of a system is given by the formula
^{1}/_{2}m_{1}u_{1}^{2}
Therefore,
the initial kinetic energy of the system is
½ x 5
x 10^{3} x 3^{2}
= 22.5
x 10^{3} J
The final
kinetic energy of the system:
K.E_{2} = K.E_{2}
(tractor) + K.E_{2} (car)
^{
}= ^{1}/_{2}m_{1}v^{2}
+ ^{1}/_{2}m_{2}v^{2} (both tractor and car
have common velocity)
= ^{1}/_{2}v^{2 }(m_{1}
+ m_{2})
= ^{1}/_{2 }x 2^{2 }(5 x 10^{3}
+ 2.5 x 10^{3})
= 2 (5 + 2.5) 10^{3}
= 2 (7.5) 10^{3}
= 15 x 10^{3} J
Therefore, loss in kinetic energy of the system = K.E_{1}
 K.E_{2}
= 22.5 x 10^{3}
 15 x 10^{3}
= 10^{3}
(22.5  15)
= 7.5
x 10^{3}
= 7.5
x 10^{3} J
