Question

A 0.4-kg cart, traveling on a horizontal air track with a speed of 6 m/s, collides with a stationary 0.8-kg cart. The carts stick together. The impulse exerted by one cart on the other has a magnitude of

A.7.2 N?s
B.1.6 N?s
C.0 N?s
D.4.8 N?s
E.2.4 N?s

Answer 1

B.1.6 N*s

Step-by-step explanation:

According to the principle of conservation of momentum, we have:

`\Delta p=0\\p_f-p_i=0\\p_f=p_i\\v_fm_f=m_1v_1+m_2v_2`

The final mass is obtained adding the masses of the two cars since they stick together after the collision. So,
`m_f=m_1+m_2`. Recall that the 0.4 kg cart collides with the stationary 0.8-kg cart. So
`v_2=0`

`v_f(m_1+m_2)=m_1v_1\\v_f=(m_1v_1)/(m_1+m_2)\\v_f=(0.4kg(6(m)/(s)))/(0.4kg+0.8kg)\\v_f=2(m)/(s)`

The magnitude of the impulse is defined as the mass multiplied by the change in the speed:

`I=m\Delta v \\I=m(v_f-v_i)\\I=0.8kg(2(m)/(s)-0)\\I=1.6N\cdot s`