Download 2004 Q6 - Loreto Balbriggan

State Examinations Commission – Physics Higher Level, 2004.
Question 6
Define (i) force, (ii) momentum. (12)
State Newton’s second law of motion.
Hence, establish the relationship: force = mass × acceleration. (15)
A pendulum bob of mass 10 g was raised to a
height of 20 cm and allowed to swing so that it
collided with a block of mass 8.0 g at rest on a
bench, as shown. The bob stopped on impact and
the block subsequently moved along the bench.
Calculate
(i) the velocity of the bob just before the collision;
(ii) the velocity of the block immediately after the collision. (18)
The block moved 2.0 m along the bench before stopping. What was the average horizontal force
exerted on the block while travelling this distance? (11)
(Acceleration due to gravity = 9.8 m s–2)
_________________________________________________________________
Define (i) force, (ii) momentum. (12)
A force is anything that causes a change in motion of a body. It is measured in Newtons.
The momentum of a body is the product of its mass and velocity. It is measured is kg.m.s -1 .
State Newton’s second law of motion.
Hence, establish the relationship: force = mass × acceleration. (15)
Newton’s second law states that if a resultant force F acts on some object, the rate of change of
momentum of the object is proportional to that resultant force, and takes place in the direction of that
force.
mv − mu
t
m (v − u )
F∝
t
F ∝ ma
F∝
F = kma
Where u is the initial and v the final velocity of the object after a time t
during which the force F was acting.
… Define 1N as that force which causes a 1kg mass to accelerate by 1ms -2
so, 1 = k. 1. 1
that is , k = 1
so , F = ma
Chris Garvey 2004
A pendulum bob of mass 10 g was raised to a
height of 20 cm and allowed to swing so that it
collided with a block of mass 8.0 g at rest on a
bench, as shown. The bob stopped on impact and
the block subsequently moved along the bench.
Calculate
(i)
the velocity of the bob just before the collision;
Loss in potential energy of bob = gain in the kinetic energy of bob
mgh = ½mv 2
gh = ½v2
2gh = v 2
√2gh = v
√(2 x 9.8 x 0.2) = v
v = 1.98ms -1
(ii)
the velocity of the block immediately after the collision. (18)
From principle of conservation of momentum, total momentum before collision = total after.
0.01 x 1.98 = 0.008 x v block
v block = 2.48 ms -1 .
The block moved 2.0 m along the bench before stopping. What was the average horizontal force
exerted on the block while travelling this distance? (11)
The deceleration of the block,
a = (2.482 – 02 )/(2 x 2.0)
a = 1.54 ms -2
… (v 2 = u 2 + 2as)
Average horizontal force F = ma
F = 0.008 x 1.54
F = 0.012N
Chris Garvey 2004