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

Answer:

B.1.6 N*s

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 $v_2=0$

$v_f(m_1+m_2)=m_1v_1\\v_f=\frac{m_1v_1}{m_1+m_2}\\v_f=\frac{0.4kg(6\frac{m}{s})}{0.4kg+0.8kg}\\v_f=2\frac{m}{s}$

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

$I=m\Delta v \\I=m(v_f-v_i)\\I=0.8kg(2\frac{m}{s}-0)\\I=1.6N\cdot s$