Question

1A.)A railroad car of mass 21700 kg moving with a speed of 3.36 m/s collides
and couples with another rail car of mass 41700 kg, moving in the same
direction with a speed of 1.34 m/s. What is the speed of the two coupled
cars after the collision?

1B.)A 0.455-kg ball traveling at 7.36 m/s west collides head on with a 1.01-kg
ball moving in the opposite direction at a velocity of 12.4 m/s. The 0.455-
kg ball rebounds at 15.4 m/s after the collision. Find the velocity of the
second ball. (use west as the positive direction)

Answer 1

Answer:

(a) 2.03 m/s

(b) 2.15 m/s

Explanation:

(a)

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

$m_1v_1+m_2v_2=(m_1+m_2)v_c$ where $v_c$ is the common velocity, $v_1$ and $v_2$ are velocities of the first railroad car  and the second railroad car respectively, $m_1$ and $m_2$ are masses of the first railroad car  and the second railroad car respectively

Substituting the given figures then

$21700\times 3.36+41700\times 1.34=(21700+41700)v_c\\63400v_c=128790\\v_c=\frac {128790}{63400}=2.031388013\approx 2.03 m/s$

(b)

Momentum, p=mv where m is the mass and v is the velocity

$m_1v_1+m_2v_2=m_1v_3+m_2v_4$ where $v_1$ and $v_2$ are initial velocities of the ball moving towards west and the other ball moving in opposite direction respectively, $m_1$ and $m_2$ are masses of the first railroad car  and the second railroad car respectively

Taking west as positive then the opposite direction will be negative hence

$0.455\times 7.36+(1.01\times -12.4)= 0.455\times -15.4 + (1.01\times v_4}$

Note that when the 0.455 rebounds, it moves towards East and since we took West as positive that is why we give 15.4 as a negative value.

$-2.1682=1.01v_4\\v_4=\frac {-2.1682}{1.01}=-2.146732673\approx 2.15 m/s$ towards East