Question

A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. When the sled crashes into the boulder, he is propelled backwards from the boulder. The collision is an elastic collision. If the boy’s mass is 40.0 kg and the sled’s mass is 2.50 kg, what is the speed of the sled and the boulder after the collision?

Answer 1

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$            (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$         (2)

Now, from  eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

Answer 2

Answer:

0.4 m/s    

Explanation:

Given:

Mass of the boy,  $m_b=40kg$

Mass of the sled, $m_s= 2.5 kg$

mass of the boulder, $M= 1000 kg$

velocity of boy + sled is before the collision, $u = 10 m/s$

After the collision, the sled and boulder stick while the boy is propelled and he continues to slide on the ice.

In elastic collision, both the momentum and kinetic energy is conserved.  

Conservation of momentum:

$(m_b+m_s)u+Mu'= m_b v+(M+m_s)v'\\ \Rightarrow (40+2.5)(10)+0=40v+(1000+2.5)v'\\ \Rightarrow 425 = 40 v+ 1002.5v' \\ \Rightarrow v= \frac{425-1002.5v'}{40}$

Conservation of kinetic energy:

$\frac{1}{2}(m_b+m_s)u^2+0=\frac{1}{2}m_bv^2+\frac{1}{2}(m_s+M)v'^2$

$\frac{1}{2}(42.5)10^2+0=\frac{1}{2}40v^2+\frac{1}{2}(1002.5)v'^2$

Substitute the value of v

$2125=20(\frac{425-1002.5v'}{40})^2+501.25v'^2$

$170000=(425-1002.5v')^2+501.25v'^2$

On solving the quadratic equation, the final velocity of the sled and the boulder is about 0.4 m/s.