Question

A 2150 kg car, moving East at 10 m/s, collided and joins with a 3250 kg car. The cars move East together at 5.22 m/s. What is the 3250 kg car’s initial velocity calculated to the nearest tenth?

Answer 1

Answer:

2.1 m/s

Explanation:

Momentum is conserved hence sum of initial momentum equals the sum of final momentum

$M_1\times v_1 + M_2\times v_2= (M_1+ M_2)\times V_c$

where $M_1$ is the mass of 2150 Kg car, $M_2$ is the mass of 3250 Kg car, $V_1$ is the velocity of the 2150 Kg car, $V_2$ is the velocity of the 3250 Kg car, $V_c$ is the common velocity

Making $V_2$ the subject then

$v_2=\frac {(M_1+ M_2)\times V_c -M_1\times v_1 }{M_2}$

Substituting 2150 Kg for $M_1$, 3250 Kg for $M_2$, 10 m/s for $v_1$, 5.22 m/s for $v_c$ then we obtain

$v_2=\frac {(2150 kg+3250 kg)\times 5.22 m/s- 2150 kg\times 10 m/s}{3250}=2.057846154 m/s\approx 2.1 m/s$