Question

A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force for 0.20 m. You then remove your hand, and the cart slides 0.35 m and strikes the 0.50-kg cart. What is the work done by you on the two-cart system? How far does the system's center of mass move while you are pushing the 1.0-kg cart? By what amount does your force change the kinetic energy of the system's center of mass?

Answer 1

Answer:

a)$W=0.8J$

b) $d_t=0.20m$

c) $\triangle K.E=0.267J$

Explanation:

From the question we are told that:

Mass of cart 1 $M_1=1.0kg$

Mass of cart 1 $M_2=0.05kg$

Force on  cart 1 $F_1=4.0N$

Push Distance of cart 1 $d_1=0.20m$

Slide Distance of cart 1 $d_1'=0.35m$

a)

Generally the equation for work-done is mathematically given by

$W=f*d\\W=4*0.20\\W=0.8J$

b)

The systems center of mass moved a net totally of (while being pushed)

Mass 1 =0.20m

Mass 2=0

Therefore

$d_t=d_1+d_2$

$d_t=0.20+0$

$d_t=0.20m$

c)

Since work-done is equal to K.E energy of cart 1

Therefore

$W=1/2mv^2$

$V_1=\sqrt{\frac{W}{1/2m}}$

$V_1=\sqrt{\frac{0.8}{1/2(1)}}$

$V_1=1.264$

Therefore Kinetic energy before collision is

$K.E_1=1/2mv^2$

$K.E_1=1/2*1*1.264^2$

$K.E_1=0.768$

Generally from the equation for conservation  of momentum the Velocity of cart 2 is mathematically given by

$v_2=\frac{m_1V_1}{m_1+m_2}$

$v_2=\frac{1*1.264}{1+0.5}$

$V_2=0.842m/s$

Therefore the final K.E is mathematically given by

$K.E_2=(1/2)(m_1+m_2)V_2^2$

$K.E_2=1/2*(1.5)(0.842)^2$

$K.E_2=0.531J$

Generally the Change in K.E is mathematically given by

$\triangle K.E=K.E_1-K.E_2$

$\triangle K.E=0.798-0.531$

$\triangle K.E=0.267J$

Therefore the will force change the kinetic energy of the system's center of mass by

$\triangle K.E=0.267J$