A tennis ball of mass m = 0.060 kg and speed v = 15 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° (Fig. 7-29). What is the impulse given to the wall?
I know the direction is normal and into the wall, but I cant get the magnitude!
impulse is the change in momentum
the velocity (parallel to the wall) is constant along its flight path
the change in velocity (and momentum) is as you say ... normal to the wall
... the incoming and outgoing speeds are both ... 15 m/s * sin(45º)
To find the impulse given to the wall, we can use the principle of conservation of linear momentum. The impulse can be determined by comparing the momentum before and after the collision.
Initially, let's consider only the horizontal component of the tennis ball's momentum. The initial momentum is given by:
p_initial = m * v_initial
Since the ball strikes the wall at a 45° angle, the horizontal component of the initial velocity can be found using trigonometry:
v_initial_horizontal = v_initial * cos(45°)
Now, let's consider the change in momentum of the tennis ball after the collision. The final momentum in the horizontal direction can be calculated in the same way:
p_final = m * v_final
Since the ball rebounds with the same speed at a 45° angle, the horizontal component of the final velocity can be found as well:
v_final_horizontal = v_final * cos(45°)
Now, the impulse imparted on the wall can be calculated as the difference between the initial and final momenta:
impulse = p_final - p_initial
In this case, the final horizontal momentum is in the opposite direction to the initial horizontal momentum. Therefore, you can consider the impulse as negative, indicating the change in direction.
So, to determine the magnitude of the impulse given to the wall, you can substitute the values of m, v, and the appropriate trigonometric functions into the equations mentioned above and calculate the impulse using the given values.
To find the impulse given to the wall by the tennis ball, we can use the impulse-momentum principle, which states that the impulse experienced by an object is equal to the change in momentum of that object.
The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of the tennis ball is given as 0.060 kg and its initial velocity is 15 m/s.
Before the collision with the wall, the ball is moving at an angle of 45° with the horizontal. However, since the angle of the rebound is also 45°, the horizontal component of its velocity remains the same after the collision. Therefore, only the vertical component of the velocity changes.
To calculate the vertical component of the initial velocity, we can use trigonometry. Since the angle of projection is 45°, the vertical component of the velocity is:
Vertical component of velocity = initial velocity * sin(angle)
= 15 m/s * sin(45°)
= 10.61 m/s
After the rebound, the vertical component of the tennis ball's velocity will now be in the opposite direction, but the same magnitude as before. Therefore, the change in velocity in the vertical direction is:
Change in velocity (Δv) = final velocity - initial velocity
= -10.61 m/s - 10.61 m/s
= -21.22 m/s
Now, we can find the impulse using the impulse-momentum principle:
Impulse = change in momentum
= mass * Δv
= 0.060 kg * (-21.22 m/s)
= -1.27 kg·m/s
Since impulse is a vector quantity, the negative sign indicates that the impulse is exerted in the opposite direction. Therefore, the impulse given to the wall by the tennis ball is 1.27 kg·m/s in the direction opposite to the rebound direction.