Question

A small, 150 g cart is moving at 1.60 m/s on a frictionless track when it collides with a larger, 5.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision?

Answer 1

Answer:

Mass of larger cart will be 0.0213 m /sec

Explanation:

We have given mass of small cart $m_1=150gram=0.150kg$

Initial velocity of small cart $v_{1i}=1.6m/sec$

Mass of larger block $m_2=5kg$

Initial velocity of larger cart $v_{2i}=0m/sec$

After the collision velocity of smaller cart $v_{1f}=-0.890m/sec$

Now according conservation of momentum

$m_1v_{1i}+m_2v_{2i}=m_1v{1f}+m_2v_{2f}$

$0.150\times 1.6+5\times 0=0.150\times -0.890+5\times v_{2f}$

$5\times v_{2f}=0.1065$

$v_{2f}=0.0213m/sec$

Answer 2

The speed of the large cart after the collision is 0.0231m/s

To get the speed of the large cart, we will use the law of conservation of momentum expressed as:

$m_1u_1+m_2u_2 = (m_1v_1+m_2v_2$

$m_1$ and $m_2$ are the masses of the object

$u_1$ and $u_2$ are the velocities of the object

$v_1 \ and \ v_2$ are the final velocity of the objects

Given the following parameters

$m_1$ = 150g = 0.15kg

$m_2$ = 5.00kg

$u_1 =1.60m/s\\u_2=0m/s (at \ rest )\\v_1 = 0.890m/s$

Substitute the given parameters into the formula to have:

$0.15(1.60)+5(0) = 0.15(0.890) + 5v_2\\0.24=0.1335 + 5v_2\\5v_2=0.24-0.1335\\5v_2=0.1065\\v_2=0.1065/5\\v_2=0.0213m/s$

Hence the speed of the large cart after the collision is 0.0231m/s

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