A ball has a mass of 2.0 kg. The ball approaches a wall at a speed of 3.0 m/s and rebounds at a
speed of 1.0m/s.
wall
What is the impulse on the wall?
A 4.0N
B 4.0NS
C 8.0N
D 8.0Ns
The Correct Answer and Explanation is :
The correct answer is:D) 8.0 Ns.
To find the impulse on the wall due to the ball’s collision, we first need to understand the relationship between impulse, momentum, and the change in velocity of the ball.
Impulse ((J)) is defined as the change in momentum of an object, which can also be expressed as the product of the average force ((F)) applied over a time interval ((t)), or as the change in momentum:
[
J = \Delta p = p_{final} – p_{initial}
]
where momentum ((p)) is given by the product of mass ((m)) and velocity ((v)):
[
p = mv
]
In this case, the mass of the ball ((m)) is (2.0 \, \text{kg}).
- Initial momentum ((p_{initial})) when the ball approaches the wall:
[
p_{initial} = m \cdot v_{initial} = 2.0 \, \text{kg} \cdot 3.0 \, \text{m/s} = 6.0 \, \text{kg m/s}
]
- Final momentum ((p_{final})) when the ball rebounds:
Since the ball rebounds in the opposite direction, we consider the final velocity to be negative:
[
p_{final} = m \cdot v_{final} = 2.0 \, \text{kg} \cdot (-1.0 \, \text{m/s}) = -2.0 \, \text{kg m/s}
]
- Change in momentum:
Now, we calculate the change in momentum:
[
\Delta p = p_{final} – p_{initial} = -2.0 \, \text{kg m/s} – 6.0 \, \text{kg m/s} = -8.0 \, \text{kg m/s}
]
The negative sign indicates that the momentum has changed direction, but impulse (as a magnitude) is simply taken as a positive value, thus:
[
|J| = 8.0 \, \text{Ns}
]
Thus, the impulse on the wall is 8.0 Ns. Therefore, the correct answer is:
D) 8.0 Ns.
In conclusion, the impulse on the wall due to the collision is equal to the change in momentum of the ball. Since the ball’s momentum changes from a positive value to a negative value upon rebounding, the magnitude of the impulse experienced by the wall is (8.0 \, \text{Ns}).