Question 14
(a)
State the principle of conservation of momentum.
[2]
(b)
Two blocks, A and B, are on a horizontal frictionless surface. The
blocks are joined together by a spring, as shown in Fig. 2.1.
Fig. 2.1
Block A has mass 4.0
kg and block B has mass 6.0 kg.
The variation of the
tension F with the extension x
of the spring is shown in Fig. 2.2.
Fig. 2.2
The two blocks are
held apart so that the spring has an extension of 8.0 cm.
(i)
Show that the elastic potential energy of the spring at an extension
of 8.0 cm is 0.48 J. [2]
(ii)
The blocks are released from rest at the same instant. When the
extension of the spring becomes zero, block A has speed vA and block B has speed vB.
For the instant when
the extension of the spring becomes zero,
1.
use conservation of momentum to show that
kinetic energy of block A / kinetic energy of block B = 1.5
[3]
2.
use the information in (b)(i) and (b)(ii)1
to determine the kinetic energy of block A. It
may be assumed that
the spring has negligible kinetic energy and that air resistance
is negligible. [2]
(iii)
The blocks are released at time t =
0.
On Fig. 2.3, sketch a
graph to show how the momentum of block A varies with time t
until the extension of the spring becomes zero.
Numerical values of
momentum and time are not required.
Fig. 2.3
[2]
[Total: 11]
Reference: Past Exam Paper – March 2017 Paper 22 Q2
Solution:
(a)
The principle of conservation of momentum states the sum
of momentum of the bodies before an interaction is equal to the sum of momentum
of the bodies after the interaction for an isolated (closed) system (where no
(resultant) external force acts).
(b)
(i)
{The elastic potential energy can be obtained
by the area under the force-extension graph.}
EPE = area under graph or ½ Fx or ½ kx2 and F = kx
{Using EPE = ½ Fx}
energy = ½ × 12.0 × 8.0
× 10-2 = 0.48 J
or
{Using
EPE = ½ kx2 where
k = F / x = 12.0 / 0.08 = 150 N m-1}
energy = ½ × 150 × (8.0 × 10-2)2 = 0.48 J
(ii)
1.
{Momentum is always conserved.
Sum of momentum before interaction = 0 as both
blocks were stationary.
Sum of momentum after interaction = 4.0 vA -
6.0 vB
Sum of momentum before = Sum of momentum after
4.0 vA - 6.0 vB = 0}
4.0 vA = 6.0 vB
EK =
½ mv2
{EK of block A = ½ × 4.0 × vA2
Since 4.0 vA = 6.0 vB,
vB = 2 vA / 3
EK of block B = ½ × 6.0 × (2vA/3)2
Ratio = EK of block A / EK of block B}
Ratio = (½ × 4.0 × vA2) / (½ × 6.0 × (2vA/3)2)
Ratio = 1.5
2.
{Energy is also conserved.
Initial energy (EPE) = 0.48 J
Total energy after interaction = EK
of A + EK of B}
0.48 = EK of A + EK of
B
{EK of A / EK of B = 1.5
So, EK of B = EK of A /
1.5}
0.48 = EK of A + (EK of
A / 1.5)
0.48 = 5/3 × EK of A
EK of
A = 0.29 (0.288) J
(iii)
curve starts from origin and has decreasing
gradient
final gradient of graph line is zero
{The block
is initially at rest. Its speed is zero. So, the momentum is also zero
initially.
Momentum
depends on velocity (p = mv).
The force in
the stretched spring is the tension. As the spring is released, this force
causes the block to acceleration. So, its velocity increases with time and so
does its momentum.
But the force
on the spring depends on the extension (F = kx). As the spring is released, the
extension decreases, and so, the force and acceleration produced also decreases
with time. That is, the increase in velocity (and momentum) decreases with
time. So, the gradient of the graph decreases. (The gradient of a momentum-time
graph actually gives the force. F = Δp / Δt.)
When the
extension of the spring becomes zero the force on block A will become zero so
that the gradient of the graph becomes zero.}
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