A 3.0-kg ball with an initial velocity of (4 + 3 ) m/s collides with a wall and rebounds with a velocity of (-4 + 3 ) m/s. What is the impulse exerted on the ball by the wall?
A)+24 N$\cdot$ s
B)(-24 N$\cdot$ s)
C)+18 N$\cdot$ s
D)(-18 N$\cdot$ s)
E)+8.0 N$\cdot$ s

Question

A 3.0-kg ball with an initial velocity of (4  + 3  ) m/s collides with a wall and rebounds with a velocity of (-4  + 3  ) m/s. What is the impulse exerted on the ball by the wall?
A)+24 N$\cdot$ s 
B)(-24 N$\cdot$ s) 
C)+18 N$\cdot$ s 
D)(-18 N$\cdot$ s) 
E)+8.0 N$\cdot$ s

Quizplus · Accepted Answer

To find the impulse exerted on the ball, we need to use the impulse-momentum theorem, which states that the impulse given to an object is equal to its change in momentum.

First, let's find the initial momentum of the ball: 
p_initial = m*v_initial 
p_initial = 3 kg*(4i + 3j) m/s 
p_initial = 12i + 9j kg m/s

Next, let's find the final momentum of the ball: 
p_final = m*v_final 
p_final = 3 kg*(-24i + 3j) m/s 
p_final = -72i + 9j kg m/s

The change in momentum is then: 
Δp = p_final - p_initial 
Δp = (-72i + 9j) - (12i + 9j) kg m/s 
Δp = -60i kg m/s

The impulse exerted on the ball by the wall is equal to the change in momentum: 
J = Δp 
J = -60i kg m/s

The magnitude of the impulse is 60 N s, but the direction is opposite to the initial direction of the ball. So the answer is (F1F1F1S1F1F1F1024 N11eb8b15_7501_faae_8952_6b638c88e2e2_TB8845_11 s).

A 30-Kg Ball with an Initial Velocity of (4 ⋅\cdot⋅ S B)(-24 N

A 30-Kg Ball with an Initial Velocity of (4 $\cdot$ S
B)(-24 N