Question

A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart

Answer 1

Answer:

Mass of the cart is 750 kg

Given:

Mass of the boy, m = 50 kg

Speed of the boy, v = 10.0 m/s

Final speed of the boy with the cart, v' = 2.5 m/s

Solution:

Initially the cart is at rest and since its on the ground, height, h = 0

Now, by the conservation of energy, mechanical energy before and after will remain conserved:

KE + PE = KE' + PE'          (1)

where

KE = Initial Kinetic energy

KE' = Final Kinetic Energy

PE = Initial Potential Energy

PE' = Final Potential Energy

We know that:

Kinetic enrgy = $\frac{1}{2}mv^{2}$

Potential energy = mgh

Since, potential energy will remain zero, thus we apply the conservation of Kinetic Energy only.

Let the mass of cart be M, thus the mass of the system, m' = 50 + M

Using eqn (1):

$\frac{1}{2}mv^{2} = \frac{1}{2}m'v^{2}$

$\frac{1}{2}\times 50\times 10^{2} = \frac{1}{2}(50 + M)\times 2.5^{2}$

$5000 = 6.25(50 + M)$

M = 750 kg

Answer 2

Answer:

the mass of the cart is 150 kg

Explanation:

given,

mass of boy(m) = 50 kg

speed of boy (v)= 10 m/s                  

initial velocity of cart (u) = 0                    

final velocity of cart(V) = 2.5 m/s              

mass of the cart(M) = ?                              

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M =$\dfrac{500}{2.5} - 50$

M = 150 Kg                                          

hence, the mass of the cart is 150 kg