a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left…

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car moves right at 0.800 m/s. find the momentum of the 155 kg car afterwards. (unit = kg*m/s) remember: right is +, left is -

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car moves right at 0.800 m/s. find the momentum of the 155 kg car afterwards. (unit = kg*m/s) remember: right is +, left is -

Answer

Explanation:

Step1: Calculate initial momentum of 282 - kg car

The formula for momentum is $p = mv$. The mass of the first car $m_1=282$ kg and its initial velocity $v_1 = 3.50$ m/s. So the initial momentum $p_{1i}=m_1v_1=282\times3.50 = 987$ kg·m/s.

Step2: Calculate initial momentum of 155 - kg car

The mass of the second car $m_2 = 155$ kg and its initial velocity $v_2=- 1.88$ m/s. So the initial momentum $p_{2i}=m_2v_2=155\times(-1.88)=-291.4$ kg·m/s.

Step3: Calculate final momentum of 282 - kg car

The mass of the first car $m_1 = 282$ kg and its final velocity $v_{1f}=0.800$ m/s. So the final momentum $p_{1f}=m_1v_{1f}=282\times0.800 = 225.6$ kg·m/s.

Step4: Use conservation of momentum

According to the law of conservation of momentum $p_{1i}+p_{2i}=p_{1f}+p_{2f}$. Rearranging for $p_{2f}$ gives $p_{2f}=p_{1i}+p_{2i}-p_{1f}$. Substitute the values: $p_{2f}=987-291.4 - 225.6$. $p_{2f}=470$ kg·m/s.

Answer:

470 kg·m/s