EXPERIMENT # 8: IMPULSE AND MOMENTUM PART I: Introduction Newton expressed what we now call his second law of motion, not as �⃗� = ?�⃗�, but in terms of the rate of change of momentum of the object ??...

EXPERIMENT # 8: IMPULSE AND MOMENTUM



PART I: Introduction

Newton expressed what we now call his second law of motion, not as �⃗� = ?�⃗�, but in terms of
the rate of change of momentum of the object ?? ??⁄ . In this more general and powerful form the
law states that when an unbalanced force acts on a body during a finite but short time interval,
the change in the object's momentum depends on the product of the force and the time interval
for which the force acts.

The quantity �⃗�?? is defined as Impulse and the relationship between the change in momentum
and the Impulse is sometimes refe
ed to as the Impulse-Momentum Theorem. It states that the
integral of the force with respect to time is equal to the change in momentum of the object.

When an unbalanced force �⃗�??? acts on an object for a time interval Δt, the momentum of the
object will change over this time interval. If ?? and ?? are the initial and final momenta at the start
and end of this time interval respectively, then Newton's second law can be written as
�⃗�???Δ? = Δ?
The product �⃗�??? Δt, when summed over several small time intervals, is the integral of �⃗�??? dt and
is defined as the Impulse ?.
? = ∫ �⃗�????? =
??
??
?? − ??
Note that impulse is a vector quantity and has the same direction as the change in momentum
vector.


---------------
Impulse, J = F∆t

Unit: Ns

F can also be expressed as rate of change of momentum.
XXXXXXXXXXF = ∆p/∆t

 F ∆t = ∆p

Impulse = ∆p (Impulse-momentum Theorem)
Unit (momentum) = kg m/s
PART II: Ballistic Pendulum

The ballistic pendulum apparatus consists of a launcher, a projectile, and a pendulum. Attached
to the swinging end of the pendulum is a catcher designed to catch the projectile and hold it. The
http:
en.wikipedia.org/wiki/Newton's_laws_of_motion
http:
en.wikipedia.org/wiki/Impulse_(physics)
projectile can be a metal ball with some mass. The catcher rests directly in front of the launcher,
ight at the exit point of the projectile (fig. 1). Length of Pendulum is 2 m at angle of 45 degrees.


Once the launch mechanism is released, the projectile enters the catcher and the arm
swings up, moving past a grooved plastic strip. A ratchet mechanism on the catcher slides
past each groove, allowing motion up but not back down. Once the pendulum stops, it
stays locked in place recording the maximum height of the swing.

Data: Table 1.

Masses (g)
Ball-I 73.4
Ball-II 110.3
Pendulum 200
Heights to which pendulum rises after collision
Ball-I
Ball-II
Velocities for the ball before collision (cm/s)
Ball-I 894.7
Ball-II

Elastic Collision: Both momentum and KE conserved.

m1v1i + m2v2i = m1v1f +m2v2f (Momentum Conservation)
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2 v2f2 (Kinetic Energy Conservation)

Where i and f represent initial and final values.


In-elastic collision: Momentum conserved, KE not conserved.

m1v1i + m2v2i = (m1 + m2) Vf (Momentum Conservation)
 Find Vf.

To find height that the pendulum raises after collision: use the conservation of
mechanical energy after the collision.

(KE +PE) initial = (KE + PE) final
½ (m1+m2) Vf2 + 0 = 0 + (m1 +m2) g h

 Find h.



PART III: Moving Block Experiment

A pendulum bob with length (L = 3m), vertically positioned is perpendicularly to a Wooden block
of mass on a plane surface. The pendulum bob was raised to an angle (θ) to the horizontal and
elease to strike the wooden block for at least twice. See the table for details. As the bob hit the
wooden block, it moved a distance of Xwb at a speed (m/s) as shown in the diagram. (Acceleration
due to gravity = 9.8 m/s2)


Table 2 : Wooden Block
Trial Mass (g)
h1 (meters) Xwb
(meters)
Initial velocity (m/s) Final velocity (m/s)
1st XXXXXXXXXX
2nd XXXXXXXXXX
Final Velocity of the Wooden Block:
Vwbf = Xwb / t
Final Velocity of the Pendulum Bob:
Vbf = [2gL(1- cosθ)]
½
Angle:
φ= h2 / L; θ = 90
0 – φ; h2 = L – h1
Moving Block Experiment
A pendulum bob with length (L = 3m), vertically positioned is perpendicularly to a Wooden
lock of mass on a plane surface. The pendulum bob was raised to an angle (θ) to the horizontal
and release to strike the wooden block for at least twice. See the table for details. As the bob hit
the wooden block, it moved a distance of Xwb at a speed (m/s) as shown in the diagram.
(Acceleration due to gravity = 9.8 m/s^2)
Table 1 for the Data from Wooden Block
Trial Mass (g) Xwb (meters) h1(meters) Initial Velocity(m/s) Final Velocity (m/s)
1st XXXXXXXXXX
2nd XXXXXXXXXX
Impulse of wooden block:
Ft = ∆p (wooden block) = Mwb (Vwbf - Vwbi)
Impulse of the ball:
F∆t = ∆p (ball) = Mb (Vbf – Vbi)
Momentum = Mass x velocity
p = mv